SATHEE: Kinematics Question 413

Kinematics Question 413

Question: A ball is dropped from the top of a tower of height 100 m and at the same time another ball is projected vertically upwards from ground with a velocity $25 m s^{- 1}$ . Then the distance from the top of the tower, at which the two balls meet it’s

Options:

A) 68.4 m

B) 48.4 m

C) 18.4 m

D) 78.4 m

Show Answer

Answer:

Correct Answer: D

Solution:

Let the two balls P and Q meet at height x m from the ground after time t s from the start. We have to find distance, $B C = (100 - x)$

For ball P $S=x m, u=25 m s^{- 1}, a = - g$ From $S = u t + \frac{1}{2} a t^{2}$

$x = 25 t - \frac{1}{2} a t^{2}$ ………. (i)

For ball Q $S= (100 - x) m, u = 0, a =g$

$∴ 100 - x = 0 + \frac{1}{2} g t^{2}$ ………. (ii)

Adding eqn. (i) and (ii), we get $100=25t or t=4s$

From eqn. (i), $x=25 \times 4 - \frac{1}{2} \times 9.8 \times {(4)}^{2} = 21.6 m$

Hence distance from the top of the tower $= (100-x)m=(100-21 .6m)=78 .4m$

Play Store

App Store

Ask SATHEE

Welcome to SATHEE !
Select from 'Menu' to explore our services, or ask SATHEE to get started. Let's embark on this journey of growth together! 🌐📚🚀🎓

I'm relatively new and can sometimes make mistakes.
If you notice any error, such as an incorrect solution, please use the thumbs down icon to aid my learning.
To begin your journey now, click on "I understand".

Kinematics Question 413

Question: A ball is dropped from the top of a tower of height 100 m and at the same time another ball is projected vertically upwards from ground with a velocity 25ms−1 . Then the distance from the top of the tower, at which the two balls meet it’s

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu

Question: A ball is dropped from the top of a tower of height 100 m and at the same time another ball is projected vertically upwards from ground with a velocity $25 m s^{- 1}$ . Then the distance from the top of the tower, at which the two balls meet it’s