Question

A cricketer can throw a ball to the maximum horizontal distance 125m how much high above the ground can the cricketer throw the same Ball?

Answer 1

Distance covered will be maximum when angle of projection is 45°

So,

R= (u² sin2∅)/g

R=125

u=√1250

u=25√2

Now,

H= (u² sin²∅)/2g

u =25√2

∅=45°

g=10

H=31.25 m

So,

Height reached will be 31.25m

Note:

Shortcut to the problem, when Ф=45°

H = R/4

So,

R=125m

H=125/4

H=31.25 m

Hope this helps !!!!!!!!!!!!!!