Question

A man throws a stone up into air at an angle θ with the horizontal.a) Draw the path of the projectile and mark directions of velocity and acceleration at the highest position. b) Derive an expression for the horizontal range reached by the stone.​

Answer 1

Let us consider a ball projected at an angle e with respect to horizontal x-axis with the initial velocity u.

The point O is called the point of projection, e is the angle of projection and OB = horizontal range. The total time taken by the particle from reaching O to B is called the time of flight.

Now,

(a). The total time of flight is

Resultant displacement is zero in Vertical direction.

Therefore, by using equation of motion

s=ut-21gt2

gt=2sine

t=g2sine

(b). The horizontal range is

Horizontal range OA = horizontal component of velocity x total flight time

R=ucosexg2usine

R=gu2sin20

(c). The maximum height is

It is the highest point of the trajectory point A. When the ball is at point A, the vertical component of the velocity will be zero.

By using equation of motion

v2=u2-2as

0=u2sin20-2gH

H=2gu2sin20

Hence, this is the required solution