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54 PE
Expert-verifiedA 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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