Question

Obtain an expression for maximum height using projectile.​

Answer 1

Answer:

derive the expressions for Maximum height, time of ascent, time of descent, time of flight in the case of vertical projection:

ux=xt⇒x=ux×t.

x=ucosθ×t⇒t=xucosθ

H=uyt+12gt2⇒H=usinθ(usinθg)−12g(usinθg)2.

H=u2sin2θg−u2sin2θ2g⇒H=u2sin2θ2g.

v=u+gt⇒vy=uy+gt.

0=usinθ−gt⇒t=usinθg.

T=t+t′⇒t′=T−t.

t′=2usinθg−usinθg⇒t′=usinθg.

Explanation:

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