Question

A 1.25 kg ball is projected vertically upward from the ground with an initial speed of 5.6 m/s. How high will it rise? What is its speed when it is at half of its maximum height? Use conservation of mechanical energy to solve this problem. Show the given, required, equations, solution and answer.

Answer 1

Explanations & calculations

Given

• m
• u

Required

• g

Equations

• 1/2mv^2
• mgh

solution

$\qquad\qquad \begin{aligned} \small E_{1p}+E_{k1}&=\small E_{p2}+E_{k2}\\ \small 0+\frac{1}{2}mu^2&=\small mgh+0\\ \small h&=\small \frac{u^2}{2g}\\ &=\small \frac{(5.6ms^{-1})^2}{2\times9.8\,ms^{-2}}\\ &=\small 1.6\,m \end{aligned}$

Answer:

$\small \text{maximum height} = 1.6\,m$