Question

2. Derive the expressions for maximum height, time of ascent, time of descent, and time of flight in oblique projection
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Answer 1

Answer:

(i) Let H be the maximum height reached by the projectile in time t1 For vertical motion,

The initial velocity = u sinθ

The final velocity = 0

Acceleration = - g

∴ using, v2 = u2 + 2as 0 = u2sin2 θ - 2gH 2gH = u2sin2 θ H = u 2 s in 2 θ 2 g u2sin2θ2g

(ii) Let t, be the time taken by the projectile to reach the maximum height H. For vertical motion, initial velocity = using θ Final velocity at the maximum height = 0 Acceleration a = - g Using the equation v = u + at1 0 = using θ - gt1 gt1 = using θ t1 = u s i n θ g usinθg Let t2 be the time of descent. But t1 = t2 i.e. time of ascent = time of descent.

∴ Time of flight T = t1 + t2 = 2t1

∴ T = 2 u s I n θ g 2usinθg

(iii) Let R be the range of the projectile in a time T. This is covered by the projectile with a constant velocity cost θ. Range = horizontal component of velocity × Time of flight i.e, R = us θ. T R = us θ. 2 u s I n θ g 2usinθg R = u2sin 2 θ g u2sin2θg