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Q.

In figure, find the horizontal velocity u (in ms-1) of a projectile so that it hits the inclined plane perpendicularly. Given H = 6.25 m. Take g=10ms-2

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answer is 5.

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Detailed Solution

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vx=ux+axt

⇒ 0=ucos⁡30∘−gsin⁡30∘t

⇒ t=u3g

Sy=uyt+12ayt2

⇒ −Hcos⁡30∘=−usin⁡30∘t −12gcos⁡30∘t2

⇒ −H32=−u2u3g−12g32u23g2

⇒ u=2gH/5=2×10×6.25/5=5ms−1

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