All Solutions
Page 75: Section Review
textbf{underline{textit{Solution}}}
$$
Thus, knowing the constant of acceleration, final velocity and the duration of motion it took the ball to reach the maximum height, we can find the initial velocity by which the ball was thrown using the following equation
$$
v_f = v_i + at
$$
Rearranging the equation in terms of initial velocity, we get
$$
v_i = v_f -a t tag{1}
$$
And, if we know the initial velocity we can find the maximum height the ball would reach, knowing the acceleration constant, the duration of motion and the initial velocity, using the following equation
$$
tag{2} d= v_i t + dfrac{1}{2} at^2
$$
Substituting equation (1) in equation (2), we get the maximum height the ball would reach in terms of acceleration and duration of motion it took the ball from the moment it was thrown till it reach the maximum height.
$$
d= left( v_f -atright)t + dfrac{1}{2} at^2
$$
$d= -at^2 + dfrac{1}{2}at^2$
And since, the acceleration is acting opposite to the direction of motion, therefor we assign a negative sign to the acceleration, thus the maximum height the ball would reach is
$$
d= at^2 – dfrac{1}{2}at^2 = dfrac{1}{2}at^2tag{3}
$$
Thus, the maximum height the ball would reach in terms of the constant of acceleration and the duration of motion is given by equation (3), let’s arrange the equation in terms of acceleration we get
$$
a= dfrac{2d}{t^2} tag{4}
$$
But before answering our question, we need first to find the relation between the maximum height and the constant of acceleration at the same velocity, and as before knowing that the final velocity is zero, we use the following equation
$$
{v_f}^2 = {v_i}^2 – 2 a d
$$
Substituting the final velocity to be zero, and rearranging in terms of constant of acceleration we get
$$
a = dfrac{{v_i}^2}{2d} tag{5}
$$
item From equation (5) as it is given that the initial velocity by which the ball was thrown is the same, and since it is given that the constant of acceleration is 1/3-rd its value on Earth.\
Therefor from equation (5) it is clear that the maximum height the ball would reach is 3 times the value of the maximum height if its value on Earth if it was throwing with the same initial velocity.\
To clarify more, the maximum height the ball would reach at Earth is
[ d_{E}= dfrac{{v_i}^2}{2a_{E}} tag{a}]
And, the maximum height it will reach on mars is
[ d_{M}= dfrac{{v_i}^2}{2a_{M} tag{b} }]
and since it is given that $a_{M}=dfrac{1}{3}a_{E}$, thus substituting in equation (b), we get
[ d_{M}= dfrac{{v_i}^2}{2left( dfrac{1}{3} a_{E} right)} = 3 left( dfrac{{v_i}^2}{2a_{E}}right) tag{c}]
And, from equation (a) we substitute in equation (c), we get
[ d_{M} = 3 d_{E}]
Thus, the maximum height the ball would reach on Mars is three time the maximum height the ball would reach on Earth.\
item From equation (4), as the constant of acceleration is always constant thus the ration to the right hand side of equation (4) must have always the same ratio, such that if the distance increased the duration of motion of flight also increases such that, the ration at the end is the same.\
Thus, as
[ a_{E} = dfrac{2d_{E}}{{t_{E}}^2}]
And,
[ a_{M} = dfrac{2d_{M}}{{t_{M}}^2}]
And given, that $a_{E} = 3 a_{M}$ and from part (a) we have $d_{M}= 3 d_{E}$, thus we substituting in the above equations and equating we get
[ 3 dfrac{2d_{M}}{{t_{M}}^2} = dfrac{2d_{E}}{{t_{E}}^2}]
And substituting by $d_{M}$ in terms of $d_{E}$, we get
[ 3 dfrac{cancel{2} times 3 d_{E}}{{t_{M}}^2}= dfrac{cancel{2} d_{E}}{{t_{E}}^2}]
[ 9 dfrac{cancel{d_{E}}}{{t_{M}}^2}= dfrac{cancel{d_{E}}}{{t_{E}}^2}]
Thus, we have
[ {t_{M}}^2 = 9 {t_{E}}^2 ]
Taking, the square root of both sides we get
[ t_{M} = 3 t_{E}]
Therefor, the duration of motion the ball took to reach the maximum height on Mars is 3 time the duration on Earth.
item The maximum height on Mars the ball would reach if thrown with the same initial velocity on Earth, would be three times the maximum height the ball would reach on Earth.
item The duration taken by the ball to reach the maximum height on Mars would be three times the time taken by the ball to reach the maximum height on Earth.
The acceleration is of a constant amount $a=9.8 mathrm{m/s^2}$ and is directed towards the ground.
So ;
$v^{2}=v_{i}^{2} 2a Delta d$ $(where$ $a = -g)$
$v=sqrt{v_{i}^{2}-2g Delta d}$
$v=sqrt{(0.0m/s)^{2} – (2)(9.80 m/s^{2})(-4.3m)}$
$v= 9.2m/s$
textit{color{#c34632} $ 9.2 $ $ m/s $}
$$
– Distance traveled by the keys: $S_{key}=4.3mathrm{~m}$
**Objective**
– Find final velocity: $v_f$
Since we are given the distance that the keys travel $S_{key}$, and we know that gravity accelerates the keys downwards with acceleration of $gapprox9.8mathrm{~dfrac{m}{s^2}}$, we need to express how velocity changes, depending only on the distance traveled and the acceleration.
$$begin{align}
v=v_0+acdot t
end{align}$$
Where $v_0$ is the initial velocity, meaning $-$ velocity at the $t=0$.
Now we don’t want our equation for velocity to depend on time, since we are not given the time, but rather the distance traveled. Let’s figure out how to do that.
$$begin{align}
S=v_0cdot t+dfrac{acdot t^2}{2}
end{align}$$
Now let’s se how can we merge equations $(2)$ and $(1)$ in order to get the velocity $v$ depending only on acceleration $a$ and distance traveled $S$.
$$begin{align}
v&=v_0+acdot t ~~ /^2 nonumber
\v^2&=v_0^2+2cdot acdot tcdot v_0+(acdot t)^2
end{align}$$
Now we can factor out the term $2cdot a$:
$$begin{align}
v^2&=v_0^2+2cdot acdot(v_0cdot t+dfrac{acdot t^2}{2})
end{align}$$
$$begin{align}
v^2&=v_0^2+2cdot acdot(v_0cdot t+dfrac{acdot t^2}{2}) nonumber
\&=v_0^2+2cdot acdot S
end{align}$$
Now, using $(5)$ we can finally solve our problem.
$$begin{aligned}
v^2&=v_0^2+2cdot acdot S nonumber
\v_f^2&=0+2cdot gcdot S
\&downarrow
\v_f&=sqrt{2cdot gcdot S}
\&=sqrt{2cdot 9.81cdot 4.3}
\&=boxed{9.18mathrm{~dfrac{m}{s^2}}}
end{aligned}$$
Of course, since the keys are dropped to the ground, the direction of final velocity is downwards.
$v_{i} = v_{f} + g t_{f}$
$v_{i}= 0.0m/s + (9.80 m/s^{2})(1.5s)$
$v_{i} = 15 m / s$
textit{color{#c34632} $ 15 $ $ m/ s$ }
$$
– Time: $t=3mathrm{~sec}$
**Objective**
– Find initial velocity $v_0$
To solve this problem, we can consider the kinematic relation between the velocity of object $v$, acceleration of the object $a$, initial velocity of the object $v_0$ and time of travel $t$:
$$begin{align}
v=v_0+acdot t
end{align}$$
Since the time it takes for ball to reach the maximum height and the time it takes for ball to drop from the maximum height to the initial position is the same, it takes $dfrac{t}{2}$ time for ball to reach the maximum height, where $t$ is the time of the whole travel (ball up and ball down).
$$begin{align}
v&=v_0+acdot t nonumber
\0&=v_0+acdot dfrac{t}{2}rightarrow v_0=-acdot dfrac{t}{2}
end{align}$$
Now we can calculate the initial velocity $v_0$.
$$begin{aligned}
v_0&=-acdot dfrac{t}{2}
\&=9.81cdot dfrac{3}{2}
\&=boxed{14.71mathrm{~dfrac{m}{s}}}
end{aligned}$$
Where we ignored the minus sign, since the acceleration is opposite to the velocity when ball is going up, hence they cancel.
So ;
$Delta d= dfrac{v_{f}^{2}-v_{i}^{2}}{-2g}$
$Delta d = dfrac{(0.0 m / s)^{2}-(15 m /s)^{2}}{(-2)(9.80 m /s^{2})}$
$Delta d =11m$
textit{color{#c34632} $ 11 $ $ m $}
$$
– Time: $t=3mathrm{~sec}$
– Initial velocity: $v_0=14.71mathrm{~dfrac{m}{s}}$
**Objective**
– Find the maximum height: $h_{max}$
Since we are given time $t$, initial velocity $v_0$ and we know that acceleration due to gravity is $gapprox 9.81mathrm{dfrac{m}{s^2}}$, we need to express the maximum height $h_{max}$ depending only on these 3 quantities.
We can also note that the maximum height is equal to the half of the whole path that ball travels $S$ (Up and down), so $h_{max}=dfrac{S}{2}$. So we need to relate $S$ to $t$, $v_0$ and $a$.
$$begin{align}
v^2=v_0^2+2cdot acdot S
end{align}$$
Where $v$ is the current velocity of object (velocity at the time $t$).
We can now note that at the maximum height velocity reaches $0$, so at $S=h_{max}$, we have $v=0$. Let’s use this in equation $(1)$.
$$begin{aligned}
v^2&=v_0^2+2cdot acdot S
\0&=v_0^2+2cdot acdot h_{max}rightarrow h_{max}=-dfrac{v_0^2}{2cdot a}
end{aligned}$$
Now we can calculate $h_{max}$, using the known $v_0$ and $a$.
$$begin{aligned}
h_{max}&=-dfrac{v_0^2}{2cdot a}
\&=dfrac{v_0^2}{2cdot g}
\&=dfrac{(14.71)^2}{2cdot 9.81}
\&approxboxed{11mathrm{~m}}
end{aligned}$$
$$
F = m cdot a rightarrow a = F / m
$$
Near the earth’s surface, this ratio is constant. If we include the Force of Gravity in Newton’s second law we get:
$$
begin{align*}
a&=dfrac{F_G}{m} \
&=dfrac{mg}{m} \
&=g
end{align*}
$$
A simple experiment that we can check has the following setup. We throw the ball in the air. At the highest point, we set a measuring scale in a short range. Let’s record a few attempts in slow motion. Then, by analyzing the images, we can determine the speed just before and just after passing through the maximum point. We plot the obtained data in a $v-t$ diagram. From this diagram we would see that the velocity changes linearly and from this slope we could calculate the acceleration which should be $g = 9.8 mathrm {m/s^2}$
