Question

A ball is projected vertically upward with speed of 50 m/s. Find a. the maximum height, b. the time to reach themaximum height, c. the speed at half the maximum height. Take g=`10 m/s^2`.

Answer 1

Correct Answer - B::C
Given, `u=50 m/s
g=-10 m/s^2`
when moving upward
` v=0` t highest point a.
`S=(v^2-u^2)/(2a)
=((0.50)^2)/)2(-10))=125m
` maximum height reached = 125m
b. `t= (v-u)/a=(0-50)/(-10)=5 sec
c. S= 125/2= 62.5 m
v=?
u= 50 m/s
a = -10 m/s^2
From v^2-u^2=2aS
v=sqrt((u^2+2as))
=sqrt((50)^2+2(-10)(62.5))
=sqrt((2500-1250))
sqrt(1250)=35 m/s`.