
A body dropped from the top of a tower fell through 60 m during the last 2s of its fall. Then find the height of the tower. (take, \[g = 10\,m{s^{ - 2}}\])
A. 95 m
B. 60 m
C. 80 m
D. 90 m
Answer
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Hint: If a body or an object falls towards earth due to the gravitational force of earth and without any other force acting on it is called free fall. As this freely falling body moves under acceleration due to gravity, therefore, its acceleration remains constant. So, it is said to be a uniformly accelerated motion.
Formula Used:
From the second equation of motion we have,
\[S = ut + \dfrac{1}{2}a{t^2}\]
Where, S is distance covered or displacement, u is initial velocity, a is acceleration, t is time of motion and \[\lambda \] is threshold wavelength.
Complete step by step solution:
Consider a body that is dropped from the top of a tower that fell through 60 m during the last 2s of its fall. Then we need to find the height of the tower. The distance covered in the last 2seconds is \[{S^1} = 60m\]. Let’s assume the total time of free fall is t seconds, and the total distance covered is $S$.
Now to calculate the total time t of the free fall given by the second equation of motion,
\[S = ut + \dfrac{1}{2}a{t^2}\]
Here, \[u = 0\], \[t = t\] and \[a = g = 10\,m{s^{ - 2}}\]
Substitute the above values in the above equation we get,
\[S = \left( 0 \right)t + \dfrac{1}{2}\left( {10} \right){t^2}\]
\[\Rightarrow S = 5{t^2}\]………… (1)
The distance covered by a body in the last 2s is,
\[{S_1} = u\left( {t - 2} \right) + \dfrac{1}{2}a{\left( {t - 2} \right)^2}\]
\[\Rightarrow {S_1} = \left( 0 \right)\left( {t - 2} \right) + \dfrac{1}{2}\left( {10} \right){\left( {t - 2} \right)^2}\]
\[\Rightarrow {S_1} = 5{\left( {t - 2} \right)^2}\]………….. (2)
Now, we can write as,
\[S - {S_1} = {S^1}\]
Substitute the value of S,\[{S_1}\] and \[S'\] in above equation, we get,
\[5{t^2} - 5{\left( {t - 2} \right)^2} = 60\]
\[\Rightarrow 5{t^2} - 5{\left( {{t^2} - 2\left( 2 \right)t + {2^2}} \right)^2} = 60\]
\[\Rightarrow 5{t^2} - 5{\left( {{t^2} - 4t + 4} \right)^2} = 60\]
\[\Rightarrow 5{t^2} - 5{t^2} + 20t - 20 = 60\]
\[\Rightarrow 20t = 80\]
\[\Rightarrow t = 4s\]
Now substitute the value of t in equation (1) we get,
\[S = 5{\left( 4 \right)^2}\]
\[\therefore S = 80\,m\]
Therefore, the height of the tower is 80m
Hence, option C is the correct answer.
Note:Here we need to remember the equation of motion for the displacement or the total distance travelled. The free fall of the body is influenced by the gravitational force acting on the body. Here the factors affecting the motion of the body in the vertical direction also need to be considered to resolve the given problem.
Formula Used:
From the second equation of motion we have,
\[S = ut + \dfrac{1}{2}a{t^2}\]
Where, S is distance covered or displacement, u is initial velocity, a is acceleration, t is time of motion and \[\lambda \] is threshold wavelength.
Complete step by step solution:
Consider a body that is dropped from the top of a tower that fell through 60 m during the last 2s of its fall. Then we need to find the height of the tower. The distance covered in the last 2seconds is \[{S^1} = 60m\]. Let’s assume the total time of free fall is t seconds, and the total distance covered is $S$.
Now to calculate the total time t of the free fall given by the second equation of motion,
\[S = ut + \dfrac{1}{2}a{t^2}\]
Here, \[u = 0\], \[t = t\] and \[a = g = 10\,m{s^{ - 2}}\]
Substitute the above values in the above equation we get,
\[S = \left( 0 \right)t + \dfrac{1}{2}\left( {10} \right){t^2}\]
\[\Rightarrow S = 5{t^2}\]………… (1)
The distance covered by a body in the last 2s is,
\[{S_1} = u\left( {t - 2} \right) + \dfrac{1}{2}a{\left( {t - 2} \right)^2}\]
\[\Rightarrow {S_1} = \left( 0 \right)\left( {t - 2} \right) + \dfrac{1}{2}\left( {10} \right){\left( {t - 2} \right)^2}\]
\[\Rightarrow {S_1} = 5{\left( {t - 2} \right)^2}\]………….. (2)
Now, we can write as,
\[S - {S_1} = {S^1}\]
Substitute the value of S,\[{S_1}\] and \[S'\] in above equation, we get,
\[5{t^2} - 5{\left( {t - 2} \right)^2} = 60\]
\[\Rightarrow 5{t^2} - 5{\left( {{t^2} - 2\left( 2 \right)t + {2^2}} \right)^2} = 60\]
\[\Rightarrow 5{t^2} - 5{\left( {{t^2} - 4t + 4} \right)^2} = 60\]
\[\Rightarrow 5{t^2} - 5{t^2} + 20t - 20 = 60\]
\[\Rightarrow 20t = 80\]
\[\Rightarrow t = 4s\]
Now substitute the value of t in equation (1) we get,
\[S = 5{\left( 4 \right)^2}\]
\[\therefore S = 80\,m\]
Therefore, the height of the tower is 80m
Hence, option C is the correct answer.
Note:Here we need to remember the equation of motion for the displacement or the total distance travelled. The free fall of the body is influenced by the gravitational force acting on the body. Here the factors affecting the motion of the body in the vertical direction also need to be considered to resolve the given problem.
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