Question

A ball is thrown up with a speed of 15 m/s. How high will it go before it begins to fall? (g = 9.8 m/s²)

Answer 1

The initial velocity of the ball, u = 15 m/s

Acceleration = $-$g

Final velocity, v = 0 m/s

Let the maximum height attained by the ball before it begins to fall back = H

Using the third equation of motion, $v^2=u^2-2gH$

$$\begin{gathered} \Rightarrow 0=(15{)}^2-2\times 9.8\times H \\ \Rightarrow 19.6H=(15{)}^2 \\ \Rightarrow H=\frac{(15{)}^2}{19.6}=\frac{225}{19.6}=11.48\mathrm{m} \\ \Rightarrow H\mathit{}=11.48\mathrm{m} \end{gathered}$$

Hence, the maximum height attained by the ball is 11.48 m.