Vertical projectile motion  (Page 2/3)

We are required to determine the maximum height reached by the ball and how long it takes to reach this height. We are given the initial velocity $v_i$ = 10 m $·$ s ${}^{-1}$ and the acceleration due to gravity g = 9,8 m $·$ s ${}^{-2}$ .

Choose down as positive. We know that at the maximum height the velocity of the ball is 0 m $·$ s ${}^{-1}$ . We therefore have the following:

• $v_i=-10\mathrm{m}·{\mathrm{s}}^{-1}$ (it is negative because we chose downwards as positive)
• $v_f=0\mathrm{m}·{\mathrm{s}}^{-1}$
• $g=+9,8\mathrm{m}·{\mathrm{s}}^{-2}$

We can use:

to solve for the height.

The value for the displacement will be negative because the displacement is upwards and we have chosen downward as positive (and upward as negative). The height will be a positive number, $h=5,10$ m.

We can use:

to solve for the time.

The ball reaches a maximum height of 5,10 m.

The ball takes 1,02 s to reach the top.

We need to find how high the ball goes. We know that it takes 10 seconds to go up and down. We do not know what the initial velocity of the ball ( $v_i$ ) is.

A problem like this one can be looked at as if there are two parts to the motion. The first is the ball going up with an initial velocity and stopping at the top (final velocity is zero). The second motion is the ball falling, its initial velocity is zero and its final velocity is unknown.

Choose down as positive. We know that at the maximum height, the velocity of the ball is 0 m $·$ s ${}^{-1}$ . We also know that the ball takes the same time to reach its maximum height as it takes to travel from its maximum height to the ground. This time is half the total time. We therefore know the following for the second motion of the ball going down:

• $t=5\mathrm{s}$ , half of the total time
• $v_{top}=v_i=0\mathrm{m}·{\mathrm{s}}^{-1}$
• $g=9,8\mathrm{m}·{\mathrm{s}}^{-2}$
• $\Delta x=$ ?

We do not know the final velocity of the ball coming down. We need to choose an equation that does not have $v_f$ in it. We can use the following equation to solve for $\Delta x$ :

In the second motion, the displacement of the ball is 122,5m downwards. This means that the height was 122,5m, h=122,5m.

The ball reaches a maximum height of 122,5 m.