Question

A stone is thrown straight upward and reaches a maximum height of 32.1 m above itslaunch point. What was the initial speed with which the stone was thrown upwards?Answer:m/s

Answer 1

Answer:

The initial speed = 25.08 m/s

Explanation:

The maximum height, H = 32.1 m

The initial speed, u = ?

The acceleration due to gravity, g = 9.8 m/s²

Write out the maximum height formula and solve for u

$H=\frac{u^2}{2g}$

Substitute H = 32.1 and g = 9.8

$\begin{gathered} 32.1=\frac{u^2}{2(9.8)} \\ \\ u^2=32.1(2)(9.81) \\ \\ u^2=629.16 \\ \\ u=\sqrt{629.16} \\ \\ u=25.08\text{ m/s} \end{gathered}$

The initial speed = 25.08 m/s