Question

A cricketer can throw a ball to a max horizontal dis of 100m how much high above the ground the cricketer throw the same ball

Answer 1

For maximum range the @ (theta) is 45°.
Now we know that

Rmax=vi²/2g
OR
vi²=Rmax×2g
vi²=100m×10m/s². (g=10m/s²)
vi²=1000m²/s²---------------------- (1)

Also we know that

H=vi²sin²@/2g

by putting the value of vi² from eq(1)

H=1000 × (sin 45)² /2×10
H=1000 × (0.707)² / 20
H=1000 × 0.5 / 20
H=500/20
H=25m-------------answer

OR simply by

we know that
Rmax = 4H
Rmax/4=H
100/4 =H
25=H
or
H=25m---------------answer


Hope you find ur answer