Question

A coin is dropped from a hot-air balloon that is 300m above the ground and rising at 10.00 m/s upward. For the coin, find (a) the maximum height reached, (b) its position and velocity 4.00 s after being released, and (c) the time before it hits the ground.

Short answer

(a) The maximum height covered by the stone is 305.099m.

(b) The velocity when the time is 4 s is 29.61 m/s and 44.705 meter from the balloon downwards

(c) The time before it hits the ground is 7.88 s

Step-by-step solution

Calculating the maximum height reached

(a) When the stone reaches its maximum height the velocity of the stone will be zero.

Hence the final velocity is zero in this case

The initial velocity is given 10 m/s upward

The gravitational acceleration is also upward 9.81 m/s²

The known value

\[\begin{array}{*{20}{l}}{U = {\rm{ }} - 10{\rm{ }}m/s}\\{g = - 9.81{\rm{ }}m/{s^2}}\\{V = {\rm{ }}0{\rm{ }}m/s}\end{array}\]

\[\begin{array}{c}{v^2} - {u^2} = 2gd\\{\left( 0 \right)^2} - {\left( { - 10} \right)^2} = 2\left( { - 9.81} \right)\left( d \right)\\19.61\,d = 100\\d = \frac{{100}}{{19.61}}\end{array}\]

\[d = 5.099\,m/s\]

Time taken to travel this distance

\(\begin{array}{l}v = u + gt\\0 = \left( { - 10} \right) + \left( { - 9.81} \right)\left( t \right)\\t = 0.981\,s\end{array}\)

Hence, the maximum height covered by the stone is 300+5.099 m =\[305.099 m\]

Calculating position and velocity at 4.00 s

(b) Here the initial velocity of the stone is given to be \[ - 10{\rm{ }}m/s\]

Than the ball goes to maximum height of \[5.099{\rm{ }}m\] in \[\;0.981{\rm{ }}s\]

Hence the distance travelled by the remaining time \[4 - 0.981{\rm{ }} = 3.019{\rm{ }}s\]

\[\begin{array}{*{20}{l}}{U = 0m/s}\\{g = 9.81{\rm{ }}m/{s^2}}\\{t = {\rm{ }}3.019}\end{array}\]

Position:

\(\begin{array}{c}d = ut + \frac{1}{2}g{t^2}\\d = \left( 0 \right)\left( {3.019} \right) + \frac{1}{2}\left( {9.81} \right){\left( {3.019} \right)^2}\\d = \,\,44.705\,m\end{array}\)

The distance of the stone after 4 s is \[44.705\] meter from the balloon downwards

Velocity:

\(\begin{array}{l}v = u + gt\\V = 0 + \left( {9.81} \right)\left( {3.019} \right)\\v = 29.61\,m/\,\,s\end{array}\)

The velocity when the time is 4 s is \[29.61 m/s\]

Calculating time before it hits the ground

(c) The time before it hits the ground:

\[\begin{array}{*{20}{l}}{U{\rm{ }} = 0}\\{D = {\rm{ }}305.09{\rm{ }}m}\\{g = 9.81{\rm{ }}m/{s^2}}\end{array}\]

\(\begin{array}{c}d = ut + \frac{1}{2}g{t^2}\\305.09 = \left( 0 \right)\left( t \right) + \frac{1}{2}\left( {9.81} \right){\left( t \right)^2}\\4.905{\left( t \right)^2} = 305.09\\{\left( t \right)^2} = \frac{{305.09}}{{4.905}}\end{array}\)

\(\begin{array}{c}{\left( t \right)^2} = 62.20\\t = 7.88\,s\end{array}\)

The time before it hits the ground is \[7.88 s\]

1. The maximum height covered by the stone is 300+5.099 m =\[305.099 m\]
2. The velocity when the time is 4 s is \[29.61 m/s\]and\[44.705\]meter from the balloon downwards
3. The time before it hits the ground is \[7.88 s\]