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Question 10

A particle P P\,P is projected from a point O O\,O with velocity Ums−1U\text{ms}^{-1}Ums−1 at an angle of θ∘ \theta^\circ\,θ∘ to the horizontal. When P P\,P has moved a horizontal distance xxx, its height above O O\,O is yyy.

a.

Show that

y=xtan⁡θ−gx22u2cos⁡2θ y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta} y=xtanθ−2u2cos2θgx2​
[4]
b.

Given that θ=45∘\theta = 45^\circθ=45∘ and that when x=6x = 6x=6, y=2y = 2y=2, find the speed of P P\,P at the point where x=6x = 6x=6 and y=2y = 2y=2.

[6]

Projectiles Questions

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