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

Step-by-step 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=(u^2)/(2g)`

Substitute H = 32.1 and g = 9.8

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

The initial speed = 25.08 m/s