Question

A stone is dropped from the top of a tower and travels 24.5m in the last second of its journey. The height of the tower is
(A) 44.1 m
(B) 49 m
(C) 78.4 m
(D) 72 m

Answer 1

Hint: When a body is dropped, its initial velocity can be considered to be zero. Since it is given that the body travels a given distance in the last second of its journey, we can say that the distance travelled by the body in total time of flight minus the distance travelled by the body in one second lesser is given. Using this condition, we can solve the question.

Complete Step by Step Answer:
Let the time taken by stone to fall through the complete height of building is T.
Then the stone fell by a distance of 24.5 m from T-1 to T time.
The distance travelled by the stone in T-1 time starting from rest can be calculated using second equation of motion
$S = ut + \dfrac{1}{2}a{t^2}$ ,
Since the particle starts from rest, its initial velocity can be considered to be zero
Using T-1 in place of t gives
${S_1} = 0(T - 1) + \dfrac{1}{2}9.8{(T - 1)^2}$,
${S_1} = 4.9{(T - 1)^2}$.
Where S1 denotes the distance travelled from zero to T-1 seconds
The distance travelled by the stone in T time starting from rest can also be calculated using second equation of motion
$S = ut + \dfrac{1}{2}a{t^2}$ ,
Since the particle starts from rest, its initial velocity can be considered to be zero again
Using T in place of t gives
${S_2} = 0(T) + \dfrac{1}{2}9.8{(T)^2}$,
${S_2} = 4.9{(T)^2}$.
Where $S_2$ denotes the distance travelled from zero to T seconds
Now, it is given that
${S_2} - {S_1} = 24.5$ .
The above equation represents the distance travelled from T-1 to T seconds
Putting the values of $S_2$ and $S_1$
$4.9{(T)^2} - 4.9{(T - 1)^2} = 24.5$,
Taking 4.9 common and taking it to RHS
\[{(T)^2} - {(T - 1)^2} = \dfrac{{24.5}}{{4.9}}\],
\[2T - 1 = 5\],
\[T = 3\].
So the particle takes time 3seconds to fall through the complete height of the building
Using the equation for $S_2$, we get
${S_2} = 4.9{(T)^2}$,
${S_2} = 4.9{(3)^2}$
${S_2} = 44.1$
So the total height of the building is 44.1 m
Therefore, the correct answer to the question is option : A

Note:The question can also be solved by remembering that when a particle is released it covers distance 5m, 15m, 25m, 35m etc in each second, considering acceleration due to gravity to be 10m/$s^2$. Similarly, the values are 4.9, 49 $\times$ 3, 4.9 $\times$ 5, 4.9 $\times$ 7 etc in each second, if acceleration due to gravity is taken as 9.8m/$s^2$. Since the particle is travelling 4.9 $\times$ 5m in the last second after being released from rest, it must have been the third second of its journey. So the total height of the building is 44.1m.