Question

Derive the relation for maximum height attained and horizontal range of a projectile fired with an angle  to the horizontal. How high above the ground can a cricketer throw a ball if he can throw the same ball to a maximum horizontal distance of 100 m?​

Answer 1

Explanation:

Maximum horizontal distance, R=100m

The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection is 45

∘

, i.e., =45

∘

.

The max horizontal range for a projection velocity v is given by the relation:

R

max

=

g

u

2

100=

g

u

2

The ball will achieve the maximum height when it is thrown vertically upward. For such motion, the final velocity v is zero at the maximum height H.

Acceleration, a=g

Using the third equation of motion:

v

2

−u

2

=−2gH

H=u

2

/2g=100/2=50m