A beanbag is thrown horizontally from a dorm room window a height h above the gr

Question

A beanbag is thrown horizontally from a dorm room window a height h above the ground. It hits the ground a horizontal distance h (the same distance h) from the dorm directly below the window from which it was thrown. Ignoring air resistance, find the direction of the beanbag's velocity just before impact.

Explanation / Answer

horizontal distance = h = 1.7 t we also know that an object will fall through a height h in a time: h=1/2 gt^2 => t=Sqrt[2h/g] now, if we take this expression for t and subsitute above, we get h=1.7 t = 1.7 sqrt[2h/g] square both sides h^2 =1.7^2 * 2 h/g divide through by h: h=1.7^2 *2 /g finally....recall the equation vf^2=v0^2 + 2 a d vf=final velocity, we want the final velocity in the y direction v0=initial velocity = 0 vertically a=acceleration = g d=distance traveled = h so we have: vf^2 = 2 g h = 2 g (1.7^2 * 2/g) = 2^2 * 1.7^2 so that vf = 2*1.7 = 3.4 we find the angle from tan(theta) = vy/vx = 3.4/1.7 = 2 theta = arc tan 2 = 63.3deg

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