If we assume no friction then both the gloves (either dropped or tossed) have same acceleration due to gravity and each will be considered to be a freely falling object. Hence, the acceleration of Zoe’s glove is equal to the acceleration of Mickey’s glove; both accelerations are $9.81 m/s^2$ downward.

In this problem, a girl on a trampoline jump straight upwards with velocity $v_text{i} = 4.5~mathrm{m/s^{2}}$. We find her velocity when she returns to the trampoline.

In this case, the displacement is 0, so the magnitude of the velocity is the same. However, the velocity is now pointing downwards, so her velocity when she returns to the trampoline must be negative of her initial velocity,
$$
boxed{ v_text{f} = -4.5~mathrm{m/s} }
$$

$$
v_text{f} = -4.5~mathrm{m/s}
$$

$$
tt{using the Velocity-Position equation $v_{f}^2=v_{i}^2+2aDelta x$ we can determine the velocity of the shell when it hit the rocks, we will choose the coordinate system origin at the point where the motion originates and the positive direction downwards $Rightarrow a=+g,Delta x=+14$:}
$$

$$
begin{align*}
v_{f}^2&=0+2gDelta x\
v_f&=sqrt{2g Delta x}\
&=sqrt{2*(9.81frac{m}{s^2})*(14m)}\
&=boxed{textcolor{#4257b2}{16.6frac{m}{s}}}
end{align*}
$$

$$
tt{$16.6frac{m}{s}$}
$$

In this problem, a volcano launches a lava bomb straight upwards with initial speed $v_text{i} = 28~mathrm{m/s}$. Taking upwards to be the positive direction we find the speed and direction of motion of the lava bomb after (a) $t = 2.0~mathrm{s}$ and (b) $t = 3.0~mathrm{s}$ after launch. We use $g = 9.81~mathrm{m/s^{2}}$.

The upward direction is positive, so the acceleration due to gravity is pointing downward (negative). The initial velocity is upward, so $v_text{i} = +28~mathrm{m/s}$. The position time equation of the lava bomb is

$$
begin{align*}
v &= v_text{i} + at \
v &= v_text{i} – gt tag{1}
end{align*}
$$

Part A.

For this part, we substitute $t = 2.0~mathrm{s}$ to equation (1).

$$
begin{align*}
v &= v_text{i} – gt \
&= +28~mathrm{m/s} – left( 9.81~mathrm{m/s^{2}} right) left( 2.0~mathrm{s} right) \
&= +8.38~mathrm{m/s} \
v &= boxed{ +8.4~mathrm{m/s} }
end{align*}
$$

The positive sign tells us that the object is moving upwards.

Part B.

For this part, we substitute $t = 3.0~mathrm{s}$ to equation (1).

$$
begin{align*}
v &= v_text{i} – gt \
&= +28~mathrm{m/s} – left( 9.81~mathrm{m/s^{2}} right) left( 3.0~mathrm{s} right) \
&= -1.43~mathrm{m/s} \
v &= boxed{ -1.4~mathrm{m/s} }
end{align*}
$$

The negative sign tells us that the object is moving downwards.

begin{enumerate}
item [(a)] $v = +8.4~mathrm{m/s}$
item [(b)] $v = -1.4~mathrm{m/s}$
end{enumerate}

In this problem, Bernardo steps off at a $x_text{i, B} = 3.0~mathrm{m}$ high diving board. At the same time, Michi jump off with velocity $v_text{i, M} = +4.2~mathrm{m/s}$ from a diving board of height $x_text{i, M} = 1.0~mathrm{m}$. We write the position-time direction of Bernardo and Michi, with upwards being the positive $x$ direction. We use $g = 9.81~mathrm{m/s^{2}}$

The position-time equation is

$$
begin{align*}
x &= x_text{i} + v_text{i}t + frac{1}{2}at^{2} tag{1}
end{align*}
$$

In this situation, since acceleration due to gravity is downwards, $a = -9.81~mathrm{m/s^{2}}$.

For Bernardo, he just steps out of the board so $v_text{i, B} = 0$. The position-time equation is

$$
begin{align*}
x_{B} &= x_text{i, B} + v_text{i, B}t + frac{1}{2}at^{2} \
&= 3.0~mathrm{m} + 0 – frac{9.81~mathrm{m/s^{2}}}{2}t^{2} \
x_{B} &= boxed{ 3.0~mathrm{m} – left( 4.905~mathrm{m/s^{2}} right)t^{2} }
end{align*}
$$

For Michi, we have

$$
begin{align*}
x_{M} &= x_text{i, M} + v_text{i, M}t + frac{1}{2}at^{2} \
&= 1.0~mathrm{m} + left( 4.2~mathrm{m/s} right)t – frac{9.81~mathrm{m/s^{2}}}{2}t^{2} \
x_{M} &= boxed{ 1.0~mathrm{m} + left( 4.2~mathrm{m/s} right)t – left( 4.905~mathrm{m/s^{2}} right)t^{2} }
end{align*}
$$