A physics student stands on the edge of a tall building 50.0m high. She throws a

Question

A physics student stands on the edge of a tall building 50.0m high. She throws a ball
with a speed of 20.0m/s at an angle of 38.0 degrees above the horizon.
a) when (what t) does the ball reach a vertical height (y) of 23.5m above the ground?
b) what is the velocity (expressed in unit vector form and its speed & direction) of the
ball just before hitting the ground?
c) how far from the edge of the building (what x) does the ball hit the ground?
d) what is the position (find x,y) of the ball when it reaches a speed of 25.0m/s?
(Hint: speed is the magnitude of the velocity vector, also be mindful of the
directions (signs) of vy)
e) how high (what y) from the top of the building does the ball reach?

Explanation / Answer

vx (speed in horizontal direction) = 22m/s (constant) uy (initial vertical speed) = 0 s (distance) = 48m a (grav rate) = 9.8m/s/s Use s = ut + (at^2)/2 to find t (time taken to fall vertically) 48 = 0 + (9.8t^2)/2 9.8t^2 = 96 t = 3.13s We know horizontal speed is constant. So find out how fast it will accelerate to vertically in 3.13s using v^2 = u^2 + 2as v^2 = 0 + 2 x 9.8 x 48 = 940.8 v (vertical) = 30.67m/s Using vertical and horizontal speeds, calculate the vector (a^2 = b^2 + c^2) v^2 = 30.67^2 + 22^2 = 1424.8 v = 37.75m/s is the speed it hits the beach. Use v(total) = v(horizontal) cos x to find angle x 37.75 = 22 cos x cos x = 0.582 x = 54.3°

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