A ball is thrown down vertically with an initial speed of from a height of . (a) What is its speed just before it strikes the ground? (b) How long does the ball take to reach the ground? What would be the answers to (c) part a and (d) part b if the ball were thrown upward from the same height and with the same initial speed? Before solving any equations, decide whether the answers to (c) and (d) should be greater than, less than, or the same as in (a) and (b).

Kinematic equations of motion are the set of equations that describe the motion of an object with constant acceleration. In this case the initial velocity and initial displacement has been given. Using these quantities, the kinematic equations can be constructed to find the velocity and time of the ball.

The expression for the kinematic equations of motion are given as follows:

$v=v_0+at$ … (i)

$v^2={v_0}^2+2ay$

… (ii)

Here, $v_0$ is the initial velocity, $t$ is the time, and $a$ is the acceleration.

When the ball is thrown vertically down from height h, the acceleration on the ball is positive, .

Therefore, from equation (ii),

$v={v_0}^2+2gh$

$v=\sqrt{{v_0}^2}+2gh$

Thus, the speed of ball just before it strikes the ground is $v=\sqrt{{v_0}^2}+2gh$.

When the ball is thrown vertically down from height h, the acceleration on the ball is positive, $a=g$.

Therefore, from equation (i),

$v=v_0+gt$

$t=\frac{-v_0+v}{g}$

Substitute the value of $v$ from part (a).

role="math" localid="1650518721505" $t=\frac{-v_0+\sqrt{{v_0}^2+2gh}}{g}$

Thus, the time taken by the ball to reach the ground is $t=\frac{-v_0+\sqrt{{v_0}^2+2gh}}{g}$.

If the ball would be thrown upward from the same height and with the same initial speed, after some time, it will return to the same height with the same initial speed. So, the speed of the ball just before it strikes the ground will be the same as in (a).

When the ball is thrown vertically up from height h, the acceleration on the ball is positive, $a=g$ and the initial velocity is negative.

Therefore, from equation (i),

$v=-v_0+gt$

$t=\frac{v_0+v}{g}$

Substitute the value of $v$ from part (a).

$t=\frac{v_0+\sqrt{{v_0}^2}+2gh}{g}$

Thus, the time taken by the ball to reach the ground is $t=\frac{v_0+\sqrt{{v_0}^2+2gh}}{g}$ which is greater than the time in part (a).