Answer to Question #230197 in Physics for Yzang

Question #230197
A ball is released from a window that is 10 meters above the ground. When
the ball leaves your hand, it is moving 8m/s at an angle 30 degrees below the horizontal.
Ignoring the effect of air resistance, how far horizontally from the window will the
ball hit the ground?
1
Expert's answer
2021-08-29T16:56:57-0400

Let's first find xx and yy components of the initial velocity of the ball:


v0x=v0cosθ=8 mscos30=6.93 ms,v_{0x}=v_0cos\theta=8\ \dfrac{m}{s}\cdot cos30^{\circ}=6.93\ \dfrac{m}{s},v0y=v0sinθ=8 mssin30=4 ms.v_{0y}=v_0sin\theta=8\ \dfrac{m}{s}\cdot sin30^{\circ}=4\ \dfrac{m}{s}.

Then, we can find the time that the ball takes to reach the ground from the kinematic equation. Let's take the upwards as the positive direction, then we get:


y=y0+v0yt12gt2,y=y_0+v_{0y}t-\dfrac{1}{2}gt^2,0=104t4.9t2,0=10-4t-4.9t^2,4.9t2+4t10=0.4.9t^2+4t-10=0.

This quadratic equation has two roots: t1=1.08t_1=1.08 and t2=1.89t_2=-1.89. Since time can't be negative the correct answer is t=1.08 st=1.08\ s.

Finally, we can find the range of the ball:


x=v0xt=6.93 ms1.08 s=7.48 m.x=v_{0x}t=6.93\ \dfrac{m}{s}\cdot1.08\ s=7.48\ m.

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