Question

A ball is projected vertically upward with a speed of 50 m/s. Find

(a) The maximum height,

(b) The time to reach the maximum beight,

(c) The speed at half the maximum geight. take g = $10m/s^2$

Answer 1

Given, u = 50 m/s

g =$-10m/s^2$

When moving upward

v = 0 at highest point

(a) S = $\frac{v^2-u^2}{2a}$

= $\frac{(0.50{)}^2}{2}.(-10)=125m$

Maximum height reached = 125 m\

(b) t = $\frac{(v-u)}{a}=\frac{0-50}{-10}=5sec$

(c) S =$\frac{125}{2}=62.5m,v=?$

u = 50 m/s, a = $-10m/s^2$

From $V^2-u^2=2aS$

V = $\sqrt{(u^2+2aS)}$

= $\sqrt{\left({50}^2\right)+2(-10)\left(62.5\right))}$

=$\sqrt{(2500-1250)}$

= $\sqrt{1250}=35m/s$