A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. calculate where and when the two stones will meet.

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

A stone is allowed to fall from the top of a tower $100 m$ high and at the same time another stone is projected vertically upwards from the ground with a velocity of $25 m / s$ . calculate where and when the two stones will meet.

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

100 m, 10 s

80 m, 5 s

80 m, 4 s

100 m, 5 s

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

A stone is allowed to fall from the top of a tower

100 m

high and at the same time another stone is projected vertically upwards from the ground with a velocity of

25 m / s

. The two stones will meet after

4 s

at a height of

80 m

from the ground.
Given data in the problem,
Height of tower,

s = 100 m

For first stone Initial velocity,

u = 0

By using the second equation,

s = ut + \frac{1}{2} {at}^{2}

= > a = g

Where,

s = distance

u = initial velocity

a = uniform acceleration

t = time

By putting all the values in the above equation,

= > x = 0 + \frac{1}{2} g t^{2}

, where

x

is the height where two stones meet.

= > x = 0 + \frac{1}{2} \times 10 {\times t}^{2}

= > x = 5 {\times t}^{2}

… (1)
Now, for second stone Initial velocity,

u = 25 m / s

By using the second equation,

s = ut + \frac{1}{2} {at}^{2}

Where acceleration

a = g

(acceleration due to gravity) Where,

s = distance

u = initial velocity

a = uniform acceleration

t = time

By putting all the values in the above equation,

\Rightarrow 100 – x = ut + \frac{1}{2} g t^{2}

\Rightarrow 100 – x = (25) \times t + \frac{1}{2} \times 10 {\times t}^{2}

\Rightarrow 100 – x = 25 \times t - 5 {\times t}^{2}

… (2)
Now adding equation 1 and 2

\Rightarrow 25 t = 100

\Rightarrow t = 4 s

putting value of t in equation 1, we get

\Rightarrow x = 5 \times 42

\Rightarrow x = 80 m

The stones will meet up at a height of

80 m

and time is

4 s

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th