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Free fall motion

From the foot of a tower 90 {~m} high, a stone is thrown up so

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From the foot of a tower $90 m$ high, a stone is thrown up so as to reach the top of the tower. Two second later another stone is dropped from the top of the tower. Find when and where two stones meet. (Take $g = 10 m / s^{2}$ )

Views: 6,154 students

Updated on: Jul 20, 2023

Book:

Physics Galaxy Mechanics Vol 1 (Ashish Arora)

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From the foot of a tower

90 m

high, a stone is thrown up so as to reach the top of the tower. Two second later another stone is dropped from the top of the tower. Find when and where two stones meet. (Take

g = 10 m / s^{2}

)

13 mins ago

Discuss this question LIVE

13 mins ago

Text solutionVerified

First stone is thrown so as to reach the top of the tower, so its initial velocity is

u = 2 g H = 2 \times 10 \times 90 = 42.5 m / s

Let us take the time

t = t_{0}

, when the two stones meet at a height

h

above the foot of the tower.
The first stone is travelled a height

h

in the duration

t_{0}

and the second stone has fallen a distance

(90 - h)

in time

(t_{0} - 2)

.
From speed equations, we have
For first stone

h = 42.5 t_{0} - \frac{1}{2} (10) t_{0}^{2}

For second stone

90 - h = \frac{1}{2} (10) (t_{0} - 2)^{2}

Adding above two equation,

22.5 t_{0} t_{0} = 70 = 3.11 s

Thus height

h

is given as

h = 42.5 (3.11) - \frac{1}{2} (10) (3.11)^{2} = 83.82 m

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Question Text	From the foot of a tower $90 m$ high, a stone is thrown up so as to reach the top of the tower. Two second later another stone is dropped from the top of the tower. Find when and where two stones meet. (Take $g = 10 m / s^{2}$ )
Updated On	Jul 20, 2023
Topic	Motion in Straight Line
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 3
Upvotes	411
Avg. Video Duration	8 min

From the foot of a tower $90 m$ high, a stone is thrown up so as to reach the top of the tower. Two second later another stone is dropped from the top of the tower. Find when and where two stones meet. (Take $g = 10 m / s^{2}$ )

Text solutionVerified

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From the foot of a tower 90 m high, a stone is thrown up so as to reach the top of the tower. Two second later another stone is dropped from the top of the tower. Find when and where two stones meet. (Take g=10 m/s2 )

Text solutionVerified

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From the foot of a tower $90 m$ high, a stone is thrown up so as to reach the top of the tower. Two second later another stone is dropped from the top of the tower. Find when and where two stones meet. (Take $g = 10 m / s^{2}$ )