please help me with this physics question

Question

A stunt woman of mass 50.0 kg can run at a top speed of 19.2 mph. A film requires that she run and jump off a cliff that sits 13.3 m above the surface of a lake. However, she must clear a shelf of rock which extends from the point directly below where she jumps to a point 9.2 m from the edge, in order to safely land in the water. A) Suppose that she jumps from the cliff at a 45 ̊ angle to the horizontal, and that air resistance is negligible. Can she clear the rocky shelf?

B) Now suppose her initial velocity is horizontal instead. Can she still make it?

Answer 1

I assume that the rock ledge is at the water level. So, basically, all we need is time in the air to reach a vertical displacement of -13.3 m in y and then use that to find the horizontal displacement (which must be greater than 9.2 m in order to make it)

y - y₀ = d_y = v_y0t - (1/2)gt² This leads to a quadratic equation for t_a when d_y= -13.3 m (upward as positive in y).

v₀ must be converted to m/s: 19.2 miles/hr (1609 m/1mile)(1h/3600s) = 8.581 m/s

v_y0=v_x0= v₀/sqrt(2) for 45° = 6.068 m/s

4.9t_a² - 6.068t_a -13.6 = 0 (Use quadratic formula to find t_a)

Range = v_x0t_a = 6.068t_a

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