Answer to Question #95901 in Mechanics | Relativity for Sarah

Question #95901
A ball is thrown horizontally from the top of
a building 120 m high. The ball strikes the
ground 63 m horizontally from the point of
release. What is the speed of the ball just before it
strikes the ground?
Answer in units of m/s.
1
Expert's answer
2019-10-07T10:47:18-0400

Let the height of the building be H=120mH = 120m and the horizontal travel distance L=63mL = 63 m.

The equations of motion are as follows:

vx(t)=v0xv_x(t) = v_{0x}, vy(t)=v0ygt=gtv_y(t) = v_{0y} - g t = - g t (no initial vertical component of velocity),

H=gt22H = \frac{g t^2}{2}, L=v0xtL = v_{0x} t .

From the last two equations, t=2Hgt =\sqrt{\frac{2 H}{g}}, v0x=Lt=Lg2Hv_{0x} = \frac{L}{t} = \frac{L \sqrt{g}}{\sqrt{2 H}}. The speed before hitting the ground is v=vx2+vy2=v0x2+(gt)2=gL22H+2gH50.2msv = \sqrt{v_x^2 + v_y^2} = \sqrt{v_{0x}^2 + (g t)^2} = \sqrt{\frac{g L^2}{2 H} + 2 g H} \approx 50.2 \frac{m}{s}.


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