Question

Two identical balls are thrown vertically upward. The second ball is thrown with an initial speed that is twice that of the first ball. How does the maximum height of the two balls compare?

Answer 1

Answer:

The maximum height of ball 2 is 4 times that of ball 1

Explanation:

We can find the maximum height of each ball by using the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

$a=g=-9.8 m/s^2$ is the acceleration of gravity (we take upward as positive direction)

s is the displacement

At the maximum height, s = h and v = 0 (the final velocity is zero), so re-arranging the equation:

$h=\frac{-u^2}{2g}$

The first ball is thrown with initial velocity $u_1$, so it reaches a maximum height of

$h_1 = -\frac{u_1^2}{2g}$ (the quantity will be positive, since g is negative)

The second ball is thrown with initial velocity

$u_2 = u_1$

so it will reach a maximum height of

$h_2 = - \frac{u_2^2}{2g}=-\frac{(2u_1)^2}{2g}=4(-\frac{u_1^2}{2g}) = 4h_1$

So, its maximum height will be 4 times the maximum height reached by ball 1.