Question

The maximum horizontal range of a projectile is 400m.The maximum height atteined by it?

Answer 1

Rmax =u²/g
400=u²/10
u²=4000m/s

Angle is 45° (At Rmax angle is 45°)

So, H=u²sin²ø/2g
H= 4000sin²45/2*10
H=100*1/√2*1/√2
H=200/2
H=100m

Answer 2

$\mathcal{\huge\red\star}{\red{\huge{Heymate}}}{\huge\red\star}$

✶⊶⊷⊶⊷⊷⊶⊷ ❍⊷⊶⊷⊶⊷⊶⊷✶

$\mathcal{\green{\huge{Answer}}}$

As the,

R tan$\theta$=4H
and also range is maximum at $\theta$=45
So
400× tan${45}$=4×H
400×1=4×H
400=4H
so
H=100 m

❤Thanks For Asking $\underline\bold{ RUPESH}$

✶⊶⊷⊶⊷⊷⊶⊷ ❍⊷⊶⊷⊶⊷⊶⊷✶

$\mathcal{\red{\huge {Hope\:it\:helps}}}$

⍖━━━━━━━━━━⍖

$\mathbb{\huge{\blue{R-K}}}$

$\mathcal{\huge{\green{Plz_-Follow_-Me}}}$