Question:
what is the velocity of projection if the horizontal range of projectile be a and the maximum height attained by it is b
Answer:
Range R= u2 sin2Θ/g
Maximum height H=u2 sin2 Θ /2g
According to question,
u2 sin2Θ/g =a ......(1) and
u2 sin2 Θ /2g =b .....(2)
Sin2Θ= 2sinΘ cosΘ
put this in eq (1) we get,
2u2 sinΘ cosΘ =ag
sinΘ= ag/ 2u2 cosΘ
Put this value of sinΘ in eq.(2), we get
b= a2 g /8u2 cos2 Θ
so, velocity of projection u= √a2 g /8b cos2 Θ
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