Question

A ball is thrown vertically upwards with a velocity of 40m/s. Find the greatest height attained.

Answer 1

We can find the maximum height attained by the ball from the kinematic equation:

$$v_f^2 = v_0^2 + 2gh,$$

here, $v_0 = 40 m/s$ is the initial velocity of the ball, $v_f = 0$ is the final velocity of the ball at the maximum height, $g = -10 m/s^2$ is the acceleration due to gravity and $h$ is the maximum height attained by the ball.

Then, we get:

$$0 = v_0^2 - 2gh,$$ $$v_0^2 = 2gh,$$ $$h = \dfrac{v_0^2}{2g} = \dfrac{(40 \dfrac{m}{s})^2}{2 \cdot 10 \dfrac{m}{s^2}} = 80 m.$$

Answer:

$h = 80 m.$