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54 PE

Expert-verifiedFound in: Page 11

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**A ball is thrown straight up. It passes a 2.00 m high**** window 7.50 m**** ****off the ground on its path up and takes 1.30s ****to go past the window. What was the ball’s initial velocity?**

The initial velocity of the ball is 14.48 m/s

The above problem is from the free-falling body topic

Here the window is at the height of 7.5m from the ground.

The height of the window is2m

The window takes the path of 1.30 second to pass the window

Here the initial velocity is not given

Final velocity is also not given

The displacement is 2 at the point

Acceleration will be 9.81 negative

\[\begin{array}{*{20}{l}}{U = ?}\\{V = ?}\\{T = 1.30{\rm{ }}s}\\{a = - 9.81{\rm{ }}m/{s^2}}\end{array}\]

\[d = 2{\rm{ }}m\]

\(\begin{array}{c}d = ut + \frac{1}{2}g{t^2}\\2 = \left( u \right)\left( {1.3} \right) + \frac{1}{2}\left( { - 9.81} \right){\left( {1.3} \right)^2}\\2 = \,\,1.3u + \left( { - 8.289} \right)\,\\1.3u = 2 + 8.289\end{array}\)

\(u = 7.914\,m/s\)

The initial velocity at the point below the window is \(u = 7.914\,m/s\)

Now the initial velocity before the window starts will be given as the final velocity.

Hence to find out the initial velocity we need to use again the equation of motion.

\[\begin{array}{*{20}{l}}{V = 7.914{\rm{ }}m/s}\\{U = ?}\\{D = 7.5{\rm{ }}m}\end{array}\]

\[\begin{array}{c}{v^2} - {u^2} = 2gd\\{\left( {7.914} \right)^2} - {\left( u \right)^2} = 2\left( { - 9.81} \right)\left( {7.5} \right)\\{u^2} = 62.63 + 147.15\\{u^2} = 209.78\end{array}\]

\[u = 14.48\,m/s\]

Hence the initial velocity of the ball is \[u = 14.48\,m/s\].

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