Question

Show that the maximum height and range of a projectile are $\frac{U^2 sin^2\theta}{2g} and \frac{U^2 sin^2\theta}{g}$ respectively where the terms have their regular meanings.

Answer 1

Hey buddy,

# Horizontal range-
It's maximum horizontal distance travelled by projectile.
Horizontal components,
s = R
ux = ucosθ
ax = 0
T = 2usinθ/g

By Newton's 2nd kinematic eqn,
s = ut + 0.5ut^2
R = ucosθ×2usinθ/g
R = (u^2/g)(2sinθ.cosθ)
R = u^2sin2θ / g

# Maximum height-
It's maximum vertical height reached by projectile.
Vertical components,
s = H
ux = usinθ
v = 0
ax = -g
T = 2usinθ/g

By Newton's 3rd kinematic eqn,
v^2 = u^2 + 2as
0 = (usinθ)^2 + 2(-g)H
H = u^2(sinθ)^2 / 2g

Hope this is helpful...

Answer 2

Maximum Height:

Maximum height is the maximum vertical displacement of a projectile .

Initial velocity =uy= u sinθ

Final velocity=vy=0

Acceleration=a=-g

Displacement =Hmax= H

using V²-u²=2as

- u² sin ²θ= 2(-g)H

u²sin²θ=2gH

H= u²sin²θ/2g

Horizontal Range R:

It is the maximum horizontal displacement of a projectile  

Initial velocity =uy= u cos t

Acceleration =a=0

Time of flight =T= 2usinθ/g

Displacement =s=R

using s=ut+1/2at2²

R=(ucosθ)T

=(ucosθ)(2usinθ)/g

=u²2sinθcosθ/g

R=u²sin2θ/g