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A ball is thrown toward a cliff of height with a speed of 31 and an angle of 60 above horizontal. It lands on?

"A ball is thrown toward a cliff of height h with a speed of 31 m/s and an angle of 60 degrees above horizontal. It lands on the edge of the cliff 3.7 laters.

a) how high is the cliff

b) what is the ball's impact speed ?


thank you"

Answers (6)

ralphie 25-02-2013
"∆y = h? V_oy = sin60(31) = 31√3/2 V_fy = 0 (pretty sure its safe to assume that it barely reached the cliff, so Vfy = 0) t = 3.7 s a = -9.8 m/s^2 ∆y = V_o * t + (1/2)at^2 ∆y = (31√3/2)(3.7) + (1/2)(-9.8)(3.7^2) ∆y = h = 32.25 m Now to find impact speed i just said V_fy = 0, so there must be some V_fx. ∆x? V_ox = 31cos60 = 31/2 V_fx = ? t = 3.7 s (same time as for y) a = 0 (In free fall there is only acceleration in the y direction, no x) V = V_o + at but since a = 0, there is nothing messing up the x directions trajectory since we are ignoring air resistance. So it will have the same velocity of: Vo = Vf = 31cos60 = 31/2 = 15.5 m/s"
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