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A ball is launched horizontally from a height of 19.6 m above the ground and it...

A ball is launched horizontally from a height of 19.6 m above the ground and it strikes the ground after moving 10.0 m horizontally. What was the initial speed of the ball?
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Answer #1

Since the ball is launched horizontally, the vertical component of the velocity will be zero.

Now for vertical direction(Y-direction),

using the kinematical equation,

s=ut+\frac{1}{2}at^2

since the vertical component of initial-velocity is zero, u=0. Also we have been given, s=19.6\,m. The acceleration in this case is the gravitational acceleration g=9.8\, m/s^2

\therefore 19.6=\frac{1}{2}\cdot 9.8\cdot t^2

\therefore t^2=4 \\\therefore t=2\,s

(we have ignored the negative root t=-2s because time cannot be negative)

Now the horizontal component of velocity is unaffected because there is no acceleration in horizontal direction or X-direction. So the initial horizontal velocity is same as final horizontal velocity. The ball travels 10 m of distance in 2 seconds. Therefore,

\boxed{u=v=\frac{10}{2}=5\,m/s}

Since there is no vertical component of velocity initially, this is the initial speed of the ball.

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