A ball is projected vertically up with a speed of 50 m/s. Find the maximum height , the time to reach the maximum height, and the speed at the maximum height `(g=10 m//s^(2)) (AS_(1))`

Explanation

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`.