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. Degrees below the horizontal

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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