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LenaWriter [7]
2 years ago
8

A tennis ball is thrown from ground level with velocity directed 30° above the horizontal. if it takes the ball 0.5 s to reach t

he top of its trajectory, what is the magnitude of the initial velocity?
Physics
1 answer:
velikii [3]2 years ago
7 0
The motion of the tennis ball on the vertical axis is an uniformly accelerated motion, with deceleration of g=-9.81 m/s^2 (gravitational acceleration).

The component of the velocity on the y-axis is given by the following law:
v_y(t) = v_{y0}+gt
At the time t=0.5 s, the ball reaches its maximum height, and when this happens, the vertical velocity is zero (because it is a parabolic motion): v_y(0.5 s)=0. Substituing into the previous equation, we find the initial value of the vertical component of the velocity:
v_{y0}=-gt=-(-9.81 m/s^2)(0.5 s)=4.9 m/s

However, this is not the final answer. In fact, the ball starts its trajectory with an angle of 30^{\circ}. This means that the vertical component of the initial velocity is
v_{y0}=v_0 sin 30^{\circ}
We found before v_{0y}=4.9 m/s, so we can substitute to find v_0, the initial speed of the ball:
v_0 =  \frac{v_{y0}}{sin 30^{\circ}}=9.81 m/s
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The free body diagram of this question is shown on the first uploaded image

From the question we are told that

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The mass center of the cyclist and the bicycle is m_c = 26 \ in  behind the front axle

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