Question

a cricket ball is thrown up with a velocity of 40 ms-1. how long will it take to reach the max height? what is the max height that it will reach?

Answer 1

Step-by-step explanation:

u=40m/s

v=0m/s

let g be 10m/s^2

10m/s^2=0-40m/s /t

therefore t=40/10=4s

s=ut+1/2at^2

s=40×4 +1/2×10×16

s=160+80

s=240m