All Solutions
Page 150: Practice Problems
78.4 = dfrac{1}{2}9.8t^2
$$
we will use newotns law of motion
$$
s = V_o t+dfrac{1}{2}at^2
$$
4.9t^2=78.4
$$
t^2 = 16
$$
then t = 4
5 x 4 = 20 m
then distance = velocity x time
V_f = 0-9.8 times 4
$$
$$
V_f = V_o +at
$$
b) distance = 20 m
c) vertical component = -39.2 m/sec
$t = sqrt{dfrac{-2y}{g}} = sqrt{dfrac{-2*(-0.6)}{9.8}} = 0.35 s$
The velocity of the giraffes must be:
$$
v_x = dfrac{x}{t} = dfrac{0.4}{0.35} = 1.1 m/s
$$
1.1 m/s
$$
– Height: $h = – 0.6 mathrm{~m}$;
– Distance: $s = 0.4 mathrm{~m}$;
**Required:**
– The velocity of the giraffes at the end of the belt $v$;
$$begin{align*}
v &= frac{s}{t} &&(1) \
t &= sqrt{ frac{-2y}{g} } &&(2)
end{align*}$$
$$begin{align*}
t &= sqrt{ frac{-2h}{g} } \
&= sqrt{ frac{-2 cdot (- 0.6 mathrm{~m})}{9.8 ,frac{text{m}}{text{s}^2}} } \
&= 0.35 mathrm{~s}
end{align*}$$
$$begin{align*}
v &= frac{s}{t} \
&= frac{ 0.4 mathrm{~m}}{0.35 mathrm{~s}} \
&= 1.1 ,frac{text{m}}{text{s}} \
&approx 1 ,frac{text{m}}{text{s}}
end{align*}$$
$$boxed{ v = 1 ,frac{text{m}}{text{s}} }$$
$t = dfrac{x}{v_x} = dfrac{0.070}{2.0} = 0.035 s$
They will fall a distance of y at the t=0.035s:
$y = (-1/2) g t^2 = (-1/2) (9.8) (0.035)^2 = -6.0e-3 m$
Thus they fall 6.0 mm.
– Speed: $v = 2 ,frac{text{m}}{text{s}}$;
– Distance: $s = 7 mathrm{~cm}$;
**Required:**
– The distance of the sundaes from the end of the counter $s$;
$$begin{align*}
v &= frac{s}{t} &&(1) \
t &= sqrt{ frac{-2y}{g} } &&(2)
end{align*}$$
$$begin{align*}
t &= frac{s}{v} \
&= frac{ 7 mathrm{~cm}}{ 2 ,frac{text{m}}{text{s}} } \
&= frac{ 0.07 mathrm{~m}}{ 2 ,frac{text{m}}{text{s}} } \
&= 0.035 mathrm{~s}
end{align*}$$
$$begin{align*}
y &= – frac{t^2 g }{2} \
&= – dfrac{(0.035 mathrm{~s})^2 cdot 9.8 ,frac{text{m}}{text{s}^2} }{2} \
&= 0.006 mathrm{~m}
end{align*}$$
$$boxed{ y = 0.006 mathrm{~m} }$$
$$
begin{align*}
v_i &= 27.0 ,mathrm{frac{m}{s}}\
theta &= 30.0^{circ}
end{align*}
$$
Given the equation
begin{align}
v_y = v_i sin{theta}
end{align}
where $v_y$ is the velocity of the ball with respect to the $y$-axis. Now through the $3^{rd}$ kinematic equation, we also know that when the ball lands, its $y$-component is $0$
begin{align}
y = v_{y}t – frac{1}{2}gt^{2} = 0
end{align}
Therefore, we can calculate for the hang time by first, isolating time $t$ from the equation above.
begin{align*}
v_{y}t – frac{1}{2}gt^{2} &= 0\
frac{1}{2}gt cdot cancel{t} &= v_{y}cancel{t}\
gt &= 2v_{y}\
t &= frac{2v_{y}}{g}\
intertext{We, then, substitute $v_i sin{theta}$ to $v_y$}
t &= frac{2(v_i sin{theta})}{g}
intertext{and now, we substitute the values,}
t &= frac{2(27.0,mathrm{frac{m}{s}})sin{(30.0^{circ})}}{9.8 ,mathrm{frac{m}{s^2}}}\
&= boxed{2.76 ,mathrm{s}}
end{align*}
$$
begin{align}
y = v_{y}t – frac{1}{2}gt^{2}
end{align}
$$
We will now solve for the maximum height $y$ by substituting the values.
$$
begin{align*}
y &= v_{y}t – frac{1}{2}gt^{2}\
&= (v_i sin{theta})t – frac{1}{2}gt^{2}\
&= (27.0,mathrm{frac{m}{s}})(1.38,mathrm{s}) sin{(30.0^{circ})} – frac{1}{2}(9.8 ,mathrm{frac{m}{s^2}})(1.38,mathrm{s})^2\
&= boxed{9.30 ,mathrm{m}}
end{align*}
$$
Given the equation
$$
begin{align}
v_x = v_i cos{theta}
end{align}
$$
where $v_x$ is the ball’s velocity with respect to the $x$-axis.
We can solve for the range using the formula
$$
begin{align}
x = v_{x}t
end{align}
$$
We will substitute $v_x$ with the $v_i cos{theta}$ and then substitute the values to obtain the value of the ball’s range.
$$
begin{align*}
x &= v_{x}t\
&= v_i cos{theta} cdot t\
&= (27.0 ,mathrm{frac{m}{s}})(2.76 ,mathrm{s}) cos{(30.0^{circ})}\
&= boxed{64.5 ,mathrm{m}}
end{align*}
$$
$textbf{b.)}$ $y = 9.30 ,mathrm{m}$
$textbf{c.)}$ $x = 64.5 ,mathrm{m}$
intertext{Data known:}
&v_i=27hspace{1mm}frac{m}{s}hspace{1mm}text{initial velocity}\
&theta=30^{circ}
intertext{Part a:}
intertext{We are looking for ball’s hang time:}
intertext{We known that horizontal velocity in this kind of motion remains constant, so we can find time using formula:}
&v_x=frac{x}{t}hspace{2cm}rightarrowhspace{2cm}
&t=frac{x}{v_x}\
intertext{We can find horizontal component of velocity if we use initial velocity and angle. And we can find horizontal distance $x$ by looking at a graph, and read the value:}
intertext{As we can see $x$ is between 50 and 60, so:}
&x=55hspace{1mm}m
intertext{Horizontal velocity:}
&v_x=v_{0x}=const\
intertext{As we can see $theta$ is angle between velocity (total velocity, sum of horizontal and vertical) and horizontal line, so horizontal component of velocity is:}
&v_x=v_icos 30\
&v_x=27cdot 0.866\
&v_x=23.382hspace{1mm}frac{m}{s}
intertext{Now we can find time:}
&t=frac{x}{v_x}\
&t=frac{55}{23.382}\
&t=2.35hspace{1mm}s
end{align*}
intertext{Part b:}
intertext{We are looking for the ball’s maximum height}
intertext{When ball is on maximum height, it’s vertical velocity is equal to zero, because gravity force is pushing the ball down.}
intertext{We can find it using formula:}
&y_{max}=v_{yi}t_m-frac{1}{2}gt_m^2
intertext{- is in formula because g gas different direction from y axis, and $v_{yi}$ is initial vertical velocity.}
intertext{IT takes $frac{1}{2}$ of time from part a for the ball to get to the maximum height, because when we look at x axis, maximum height is at half of total x, so time is: }
&t_m=frac{2.35}{2}\
&t_m=1.175hspace{1mm}s
intertext{We can find initial vertical velocity by using total initial velocity and angle: }
&v_{yi}=v_isin 30\
&v_{yi}=27cdot0.5\
&v_{yi}=13.5hspace{1mm}frac{m}{s}\
intertext{Now we can find $y_{max}$}
&y_{max}=13.5cdot1.175-frac{1}{2}9.8cdot1.175^2\
&y_{max}=9hspace{1mm}m
end{align*}
intertext{Part c:}
intertext{We are looking for the ball’s range.}
intertext{Range is value of x, it is how far ball get on horizontal axis, which is x axis, so:}
&R=55hspace{1mm}m
end{align*}
intertext{Part a:}
&t=2.35hspace{1mm}s
intertext{Part b:}
&y_{max}=9hspace{1mm}m
intertext{Part c:}
&R=55hspace{1mm}m
end{align*}
$$
begin{align*}
v_i &= 27.0 ,mathrm{frac{m}{s}}\
theta &= 60.0^{circ}
end{align*}
$$
Using the same method from the previous problem, we calculate the following: hang time, maximum height, and range or distance.
We know that $v_y = v_{i} sin{theta}$ such that $v_y$ is the velocity with respect to the $y$-axis. Using the equation
begin{align}
y = v_{y}t – frac{1}{2}gt^{2} = 0
end{align}
We isolate time $t$ and substitute the given values to obtain the hang time.
begin{align*}
v_{y}t – frac{1}{2}gt^{2} &= 0\
frac{1}{2}gt^2 &= v_{y}t\
gt &= 2v_{y}\
t &= frac{2v_{y}}{g}\
intertext{We substitute the given values.}
t &= frac{2(27.0,mathrm{frac{m}{s}})}{9.8,mathrm{frac{m}{s^2}}}\
t &= boxed{4.77 ,mathrm{s}}
end{align*}
Using the equation
begin{align}
y = v_{y}t – frac{1}{2}gt^{2}
end{align}
We can compute for the maximum height $y$, knowing that the ball will reach maximum height at $t_{max height} = frac{1}{2}t_{hang time} = 2.38,mathrm{s}$.
begin{align*}
y &= v_{y}t – frac{1}{2}gt^{2}
intertext{We substitute $v_y$ with $v_{i} sin{theta}$}
y &= (v_{i} sin{theta})t – frac{1}{2}gt^{2}\
&= (27.0,mathrm{frac{m}{s}})(2.38 ,mathrm{s})sin{(60.0^{circ})} – frac{1}{2}(9.8,mathrm{frac{m}{s^2}})(2.38,mathrm{s})^2\
&= boxed{27.9 ,mathrm{m}}
end{align*}
We can calculate the range $x$ of the ball given that $v_x = v_{i} cos{theta}$, such that $v_x$ is the initial velocity with respect to the $x$-axis. We will use the time $t$ which is the hang time of the ball.
begin{align}
x = v_{x}t
end{align}
begin{align*}
intertext{We substitute $v_x$ with $v_{i} cos{theta}$ and plug in the given values.}
x &= (v_{i} cos{theta})t \
&= (27.0,mathrm{frac{m}{s}})(4.77 ,mathrm{s})cos{(60.0^{circ})}\
&= boxed{64.4 ,mathrm{m}}
end{align*}
$textbf{b.)}$ Maximum height $y = 27.9 ,mathrm{m}$
$textbf{c.)}$ Range $x = 64.4 ,mathrm{m}$
intertext{Data known:}
&v_i=7hspace{1mm}frac{m}{s}hspace{1mm}text{initial velocity}\
&theta=53^{circ}hspace{1mm}text{° above the horizontal}\
&y_i=50hspace{1mm}m
end{align*}
intertext{We are looking for velocity vector, when rock hits the ground. }
intertext{Velocity has two components, horizontal component $v_x$ and vertical component $v_y$, total velocity is:}
&v^2=v_x^2+v_y^2\
&v=sqrt{v_x^2+v_y^2}\
intertext{We can solve this problem by finding those components first. We can say that motion like this can be observed as combination of two motions, motion due horizontal axis, and due to vertical. }
intertext{Let’s find horizontal component firs. We known that horizontal component of motion remains the same all the time in motion like this, so:}
&v_x=v_{xi}=const\
intertext{As we can see $theta$ is angle between initial velocity (total velocity, sum of horizontal and vertical) and horizontal line, so horizontal component of velocity is:}
&v_x=v_icos 53\
&v_x=7cdot 0.6\
&v_x=4.2hspace{1mm}frac{m}{s}
end{align*}
intertext{Now we should find vertical component of velocity. We can use formula:} &v_y^2=v_{yi}^2-2g(y-y_i)
intertext{- is in formula because g has opposite direction comparing to y axis.}
intertext{y is equal to zero, because it hits the ground, and it is 0 m.And we can fin initial vertical velocity by using angle and initial total velocity}
&v_{yi}=v_isin 53\
&v_{yi}=7cdot0.8\
&v_{yi}=5.6hspace{1mm}frac{m}{s}\
\
&v_y^2=5.6^2-2cdot9.8(0-50)\
&v_y^2=31.36+980\
&v_y^2=1011.36
intertext{value of vertical component can be positive or negative, we use negative because direction of that component when rock hits the ground is opposite comparing to direction of y axis.}
&v_y^2=sqrt{1011.36}\
&v_y=-31.8hspace{1mm}frac{m}{s}
end{align*}
intertext{Now we can find velocity:}
&v=sqrt{v_x^2+v_y^2}\
&v=sqrt{4.2^2+(-31.8)^2}\
&v=sqrt{17.64+1011.36}\
&v=32.07hspace{1mm}frac{m}{s}
end{align*}
begin{align*}
&v_x=4.2hspace{1mm}frac{m}{s}\
&v_y=-31.8hspace{1mm}frac{m}{s}\
&v=32.07hspace{1mm}frac{m}{s}
end{align*}
$$
$$
begin{align*}
y &= 50.0 ,mathrm{m}\
v_i &= 7.0 ,mathrm{frac{m}{s}}\
theta &= 53.0^{circ}
end{align*}
$$
such that $y$ is the height of the cliff, $v_i$ is the initial velocity, and $theta$ is the angle of the projectile.
$$
begin{align}
v_{ix} &= v_i cos{theta}\
v_{iy} &= v_i sin{theta} + gt
end{align}
$$
such that $v_{ix}$ is the initial horizontal velocity and $v_{iy}$ is the initial vertical velocity.
$$
begin{align}
y = v_{y}t – frac{1}{2}gt^2
end{align}
$$
We know that at $y = -50.0 ,mathrm{m}$, which is the height of the rock as it reaches the bottom of the cliff, $v_{y} = 0$, such that $v_{y}$ is the vertical velocity. therefore
$$
begin{align*}
y &= -frac{1}{2}gt^2\
-2y &= gt^{2}\
-frac{2y}{g} &= t^2\
sqrt{-frac{2y}{g}} &= t
end{align*}
$$
$$
begin{align*}
v_y &= v_1 sin{theta} + gt\
&= v_1 sin{theta} + g left(sqrt{-frac{2y}{g}} right)\
&= v_1 sin{theta} + left(sqrt{-2yg} right)
end{align*}
$$
begin{align*}
v &= sqrt{left(v_i cos{theta}right)^{2} + left(v_1 sin{theta} + left(sqrt{-2yg} right)right)^{2}}
intertext{Now, to solve for the magnitude $v$, we can plug in the given values}
v &= sqrt{left((7.0 ,mathrm{frac{m}{s}}) cos{(53.0^{circ})}right)^{2} + left((7.0 ,mathrm{frac{m}{s}}) sin{53.0^{circ}} + left(sqrt{-2(-50.0 ,mathrm{m})(9.8 ,mathrm{frac{m}{s^2}}} right)right)^{2}}\
&= boxed{37.1 ,mathrm{frac{m}{s}}}
end{align*}
$$
begin{align}
tan{theta} = frac{v_{iy}}{v_{ix}}
end{align}
$$
$$
begin{align*}
tan{theta} &= frac{v_{iy}}{v_{ix}}\
theta &= tan^{-1}{left(frac{v_{iy}}{v_{ix}}right)}\
&= tan^{-1}{left(frac{v_1 sin{theta} + left(sqrt{2yg} right)}{v_i cos{theta}}right)}\
&= tan^{-1}{left(frac{(7.0 ,mathrm{frac{m}{s}}) sin{53.0^{circ}} + left(sqrt{2(50.0 ,mathrm{m})(9.8 ,mathrm{frac{m}{s^2}}} right)}{(7.0 ,mathrm{frac{m}{s}}) cos{(53.0^{circ})}}right)}\
&= boxed{83.5^{circ}}text{ from the $+x$-axis}
end{align*}
$$
v = 37.1 ,mathrm{frac{m}{s}} text{ at } 83.5^{circ}text{ from the $+x$-axis}
$$
