Question

If $\theta$ is the angle of projection, R is the range, h is the maximum height, T is the time of flight, then show that (a) $tan \theta= \frac{4h}{R}$ and (b)$h= \frac{gT^2}{8}$.

Answer 1

Hii dear,

# Formulas-
R = u^2sin2θ / g
H = u^2(sinθ)^2 / 2g
T = 2usinθ / g

# Proofs-
a) H & R
H = u^2(sinθ)^2 / 2g
H = u^2sinθ.sinθ / 2g
sinθ = 2Hg/(u^2.sinθ) ...(1)
R = u^2sin2θ / g
R = 2u^2.sinθ.cosθ / g
cosθ = Rg/(2u^2.sinθ) ...(2)
tanθ = sinθ/cosθ
tanθ = [2Hg/(u^2.sinθ)]/[Rg/(2u^2.sinθ)]
tanθ = 4H/R

b) H & T
H = u^2(sinθ)^2 / 2g
H = (4g/g)[u^2(sinθ)^2 / g^2]
H = (g/8)[4u^2(sinθ)^2 / g^2]
H = (g/8)[(2usinθ/g)^2]
H = gT^2/8


Hope you got it...

Answer 2

a) h=u²sin²θ/2g

R= u²sin²2θ/g

h/R=u²sin²θx g/2g  x u²  2sin θcosθ

=  Tanθ/4

∴Tanθ=4h/R

B)h=u²sin²θ/2g

T=2usinθ/g

now : h/T²= u²sin²θxg²/2g

x4u²sin²θ

h/T²=g/8

∴h=gT²/8