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If four seconds be the time in which a projectile reaches a point P of its path and 5 seconds the time from P till it reaches the horizontal plane through the point of projection. The height of point P above the horizontal plane is

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100m

50m

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200m

answer is B.

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Detailed Solution

Total time of flight = 4 + 5 = 9s

$\frac{2 u \sin θ}{g} = 9 u \sin θ = 45 m / s$

Let h is the height of P above horizontal plane

h = $u \sin θ t - \frac{1}{2} g t^{2} = 45 \times 4 - \frac{1}{2} \times 10 \times 16 = 180 - 80 = 100$

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