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Question

A stone is dropped from a tower at a height of h. Another stone is thrown up simultaneously from the ground, which can reach a maximum height of 4h. Determine the time at which both the stones meet.


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Solution

Step 1: Given data

h1=h

h2=4h

u1=0

Step 2: To find

Time at which both the stones meet.

Step 3: Calculate the time of the meeting:

Use the second equation of motion to find the height, h.

s=ut+12at2

Let the initial velocity of the stone that is thrown from the ground u2.

At maximum height 4h, Velocity v2=0

v22=u22-2gs02=u22-2g4hu2=8gh

Let h1be the distance covered by the stone which is dropped from height h and h2 be the distance covered by the stone which is thrown up from the ground.

Let time t both stones crossed each other.

h1=12gt2...(1)u1=0

h2=8ght-12gt2h2=8ght-12gt2...(2)

Adding h1​ and h2 gives the height h.

Add equations (1) and (2).

h1+h2=8ghth=8ghtt=h8g

Hence, the time at which both the stones meet is h8g.


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