Question

A stone is dropped from the top of a tower and after one second, another stone is

dropped from a point 20 m below the top. If both the stones reach the ground at the

same time, calculate the height of the tower (take g = 10 m/s).​

Answer 1

Answer:

Formula used:

Equation of motion, s=ut+12at2 where s is displacement, u is initial velocity, t is time, a is acceleration.

Complete step by step answer:

Let us consider the height of the tower to be ht.

From the given question, we know that the height of the balcony is hb=ht−20m and the time difference when the both stones are thrown is Δt=1s.

The time difference when the both stones are thrown is,

Δt=t1−t2⇒1=t1−t2⇒t2=t1−1

We know that when both the stones are dropped, their initial velocity is equal to zero, u=0.

Since the stone is moving downwards the acceleration due to gravity will be taken as g=+10m/s2.

Caste I: When the stone is dropped from the tower:

Let us use the equation of motion,

s=ut+12gt2

Substitute the expression in the above equation, we have,

(0−ht)=0+12gt21

ht=12gt21 ... (I)

Caste II: When the stone is dropped from the balcony:

Similarly, from the equation of motion,

hb=0+12gt22

(ht−20)=0+12g(t1−1)2... (II)

Now we subtract equation (I) and (II),

20=12gt21−12g(t1−1)2⇒20=12g(t21−(t1−1)2)⇒40=10×(2t−1)

Simplifying the above equation,

40=10(2t1−1)⇒t1=2.5s

Now substitute the above value in equation (I), we get,

ht=12×10×(2.5)2∴ht=31.25m

Thus, the height of the tower is 31.25 m and option (C) is correct.

Note:Make sure the height of the balcony is taken into account its height is 20 m below the height of the tower. Do not use 20 m as the height of the balcony. Someone has the stone at the top of the tower to drop, so the initial velocity is taken as zero in this type of question. If you are considering the height of the tower or building as positive, then the datum is its stop.

Explanation: