Question

With what speed must a ball be thrown vertically up in order to rise to a maximum height of 45m? And for how long will it be in air?

Answer 1

1) 29.7 m/s

We can solve the first part of the problem by using the law of conservation of energy. In fact, the initial kinetic energy when the ball is thrown from the ground will be entirely converted into gravitational potential energy as the ball reaches its maximum heigth, so we can write:

$\frac{1}{2}mv^2 = mgh$

where

m is the mass of the ball

v is the initial speed

g is the acceleration of gravity

h is the maximum height

Re-arranging the equation:

$v=\sqrt{2gh}$

and substituting:

g = 9.8 m/s^2

h = 45 m

We find

$v=\sqrt{2(9.8)(45)}=29.7 m/s$

2) 6.0 s

First of all, we can calculate how long the ball takes to reach the maximum height. This can be done by using the equation:

$v=u+at$

where

v = 0 is the final speed of the ball, when it is at its maximum height

u = 29.7 m/s is the initial speed

a = -g = -9.8 m/s^2 is the acceleration of gravity (negative because it is downward)

t is the time

Solving for t,

$t= \frac{v-u}{a}=\frac{0-29.7}{-9.8}=3.0 s$

So the ball takes 3.0 s to reach the maximum heigth. The time it takes to fall down again is equal, so also 3.0 s, so the total time of the motion is

t = 3.0 + 3.0 = 6.0 s