91Ó°ÊÓ

(a) The initial height of ball is \[1.50{\rm{ }}m\], it rebound up to height of \[1.1{\rm{ m}}\]

Here the ball is free falling object.

Hence the initial velocity of the ball will be zero

The acceleration will be also 9.81 m/s²

The known value

\[\begin{array}{l}\begin{array}{*{20}{l}}{U = {\rm{ }}0{\rm{ }}m/s}\\{g = 9.81{\rm{ }}m/{s^2}}\end{array}\\d = {\rm{ }}1.5{\rm{ }}m\end{array}\]

velocity just before it strikes the floor

\[\begin{array}{c}{v^2} - {u^2} = 2gd\\{\left( v \right)^2} - {\left( 0 \right)^2} = 2\left( {9.81} \right)\left( {1.5} \right)\\{v^2} = 29.43\\v = \pm 5.42\,m/s\end{array}\]

\[V = + 5.42\,m/s\]

The final velocity of the ball when before it just strikes the ground is \[.5.42 m/s\]

(b) velocity just after it leaves the floor on its way back up

Here in this case the final velocity will be zero

The acceleration will be negative as the body is travelling to the upward direction

\[\begin{array}{*{20}{l}}{V = {\rm{ }}0{\rm{ }}m/s}\\{g = - 9.81{\rm{ }}m/{s^2}}\\{d = {\rm{ }}1.1{\rm{ }}m}\end{array}\]

\[\begin{array}{c}{v^2} - {u^2} = 2gd\\{\left( 0 \right)^2} - {\left( u \right)^2} = 2\left( { - 9.81} \right)\left( {1.1} \right)\\{u^2} = 28.449\\u = \pm 4.645\,m/s\end{array}\]

\[u = - 4.645\,m/s\]

Hence the velocity after the ball just leaves the ground surface is\[ - 4.645\,m/s\]. here the negative sign represents the upward motion.

(c) During the contact point the initial velocity is \[5.42{\rm{ }}m/s\]

The final velocity just after the contact is \[ - 4.645{\rm{ }}m/s\]

The time taken in this phenomenon is \[3.5{\rm{ }}mS\]

\[\begin{array}{*{20}{l}}{U = {\rm{ }}5.42{\rm{ }}m/s}\\{V = {\rm{ }} - 4.645{\rm{ }}m/s}\\{T = {\rm{ }}3.5{\rm{ }}x{\rm{ }}{{10}^{ - 3}}s}\end{array}\]

\(\)\(\begin{array}{l}a = \frac{{v - u}}{t}\\a = \frac{{ - 4.645 - 5.42}}{{3.5 \times 1{0^{ - 3}}}}\\a = - 2.87 \times 1{0^3}m/{s^2}\end{array}\)

Acceleration during contact with the floor if that contact lasts \[3.5{\rm{ }}ms\] is\[ - 2.87{\rm{ }}x{10^3}m/{s^2}\] . Here the negative sign represents the upward motion

Here the displacement is asked indirectly

Hence for this case when the ball is compressed

The time taken in this phenomenon is \[3.5{\rm{ }}ms\]

\[\begin{array}{*{20}{l}}{U = {\rm{ }}5.42{\rm{ }}m/s}\\{V = {\rm{ }} - 4.645{\rm{ }}m/s}\\{T = {\rm{ }}3.5{\rm{ }}x{\rm{ }}{{10}^{ - 3}}s}\\{a = - 2.87{\rm{ }}x{{10}^3}m/{s^2}}\end{array}\]

\[\begin{array}{c}{v^2} - {u^2} = 2gd\\{\left( { - 4.645} \right)^2} - {\left( {5.42} \right)^2} = 2\left( { - 2.87 \times 1{0^3}} \right)\left( d \right)\\ - 5.74 \times 1{0^3}d = - 7.8\\d = \frac{{7.8}}{{5.74 \times 1{0^3}}}\end{array}\]

\[d = 1.35 \times 1{0^{ - 3}}\,\]

Hence the ball is compressed for\[1.35 \times 1{0^{ - 3}}\,m\].

1. final velocity of the ball when before it just strikes the ground is \[5.42 m/s\]
2. The velocity after the ball just leaves the ground surface is -\[4.645 m/s\]
3. Acceleration during contact with the floor if that contact lasts 3.5 ms is\[ - 2.87 x1{0^3}m/{s^2}\]
4. The ball is compressed for\[1.35x 1{0^{ - 3}}meter\].