Question

A ball is dropped from a tower. In the last second of its motion, it travels a distance of \[15\,m\] . Find the height of the tower. \[\left[ {{\text{take}}\,\,g = 10m/{{\sec }^2}} \right]\].

Answer 1

Hint: Use the kinematic equation to determine the distance traversed by the ball before the last second and the total distance traversed by the ball. The difference in this height is given. On obtaining the total time taken by the ball, use the kinematic equation to determine the height of the tower.
Formula used:
\[S = ut + \dfrac{1}{2}a{t^2}\]
Here, u is the initial velocity, t is the elapsed time, and a is the acceleration.

Complete step by step answer:
Use the kinematic equation relating displacement, initial velocity, acceleration and time as follows,\[S = ut + \dfrac{1}{2}a{t^2}\]
Suppose the height of the tower before the last second is \[h\left( t \right)\] and the total height of the tower is \[h\left( {t + 1} \right)\].
We have given the height of the tower the last second is 15 m.
Therefore,
\[h\left( {t + 1} \right) - h\left( t \right) = 15\]
Now, use the above kinematic equation in the vertical direction by substituting \[g\] for \[a\]. Also, the initial velocity of the ball is zero.
\[\dfrac{1}{2}g{\left( {t + 1} \right)^2} - \dfrac{1}{2}g{t^2} = 15\]
\[ \Rightarrow \dfrac{1}{2}g\left( {{t^2} + 2t + 1} \right) - \dfrac{1}{2}g{t^2} = 15\]
\[ \Rightarrow gt + \dfrac{g}{2} = 15\]
\[ \Rightarrow t = \dfrac{{15 - \dfrac{g}{2}}}{g}\]
Substitute \[10\,m/{\sec ^2}\] for g in the above equation.
\[ \Rightarrow t = \dfrac{{15 - \left( {\dfrac{{10\,m/{{\sec }^2}}}{2}} \right)}}{{10\,m/{{\sec }^2}}}\]
\[\therefore t = 1\,\sec \]
Therefore, the total time is,
\[t + 1 = 2\,{\text{second}}\]
Now, we can use kinematic equation relating displacement, initial velocity, acceleration and time to calculate the height of the tower as follows,
\[h = ut + \dfrac{1}{2}g{t^2}\]
Substitute \[0\,m/\sec \] for u, 2 s for t, and \[10\,m/{\sec ^2}\] for g in the above equation.
\[h = \dfrac{1}{2}\left( {10\,m/{{\sec }^2}} \right){\left( {2\,\sec } \right)^2}\]
\[\therefore h = 20\,m\]
Therefore, the height of the tower is 20 meters.

Note:
You don’t need to take the value of g as negative as the ball is also moving in the downward direction that makes the height also negative. You can also solve this question by using the equation of time of flight in the vertical direction, \[t = \sqrt {\dfrac{{2h}}{g}} \]. Subtract 1 second from this time and substitute in the kinematic equation for calculating the vertical distance.