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Question

A tiger leaps horizontally from a 6.5-m-high rock with a speed of 3.5 m/s. How far from the base of the rock will she land?

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Answered 2 years ago
Answered 2 years ago
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x=x0+vxtx=x_0 + v_xt

y=y0+vyt12gt2y=y_0 + v_yt-\dfrac{1}{2}gt^2

These are the two projectile motion equations. We'll set the base of the cliff to be at the origin so x0=0x_0 = 0 and y0=6.5y_0 = 6.5. Also, the initial speed is vx=3.5v_x=3.5 and vy=0v_y=0

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