Question

Show that the maximum height reached by a projectile launched at an angle of 45° is one quarter of its range.

Answer 1

As we know that the formula for Hmax =V²sin²θ/2g

and the formula for range=V²sin2θ/g

first by putting the value of θ=45°,we get

Hmax=V²×sin²45°/2g                                     (value of sin45°=1/√2 )

we get Hmax =v²/4g

and similarly for range we put θ=45°

RANGE =V²sin2 ×45°/g=V²/G                  (sin2×45°=sin90°=1)

then dividing Hmac by range we get,

Hmax/range=V²/4g÷V²/g=1/4

hence Hmax=range/4

Answer 2

Hii friend,

# Given-
θ = 45°

# To prove-
H = R/4

# Proof-
Maximum Height of the projectile is given by -
H = u^2.(sinθ)^2 / 2g
H = u^2.(1/√2)^2 / 2g
H = u^2/4g ...(1)

Range of projectile is given by-
R = u^2.sin2θ / g
R = u^2.(1)/g
R = u^2/g

Putting this in eqn (1),
H = R/4

Hence proved.

Hope that was helpful..