
A metal ball falls from a height of \[32{\text{ }}m\] on a steel plate. If the coefficient of restitution is \[0.5\;\] to what height will the ball rise after a second bounce?
A. ${\text{2 }}m$
B. ${\text{4 }}m$
C. ${\text{8 }}m$
D. ${\text{16 }}m$
Answer
164.1k+ views
Hint: In this question, we have a metal ball that is dropped from a height of \[32{\text{ }}m\]and the ball reaches on the steel plate and rebounds. We need to find the rise (height) of the steel ball when it rebounds. We’ll use the formula for velocity of a body falling from a height $H$ and a relation between coefficient of restitution and velocity. Substituting all the given values in these formulas we will get the required solution.
Formula used:
If the ball is falling from the height $H$, then its velocity is;
$v = \sqrt {2gH} $
Relation between velocity and coefficient of restitution –
${v_f} = ev$
Complete step by step solution:
Given that,
Steel ball dropped from a height \[32{\text{ }}m\].
Let, the height be $H = 32{\text{ }}m$
Now, the ball reaches the plate and bounces from the steel plate. Let the height to which the steel ball rebounds be $h$. As we know, If the ball is falling from the height $H$, then its velocity is
$v = \sqrt {2gH} $
Put $H = 32{\text{ }}m$
Velocity of steel ball when it strikes the plate will be $v = \sqrt {64g} $
After striking the plate, the steel ball rebounds with the velocity ${v_f}$.
Where, ${v_f} = ev$ ----------(1)
Here, $e = $ coefficient of restitution and $v = $ velocity of striking the plate
Also, final velocity will be ${v_f} = \sqrt {2gh} $ ---------(2)
From equation (1) and (2), we get
$\sqrt {2gh} = ev$
It can be written as $\sqrt {2gh} = e\sqrt {64g} $
Squaring both the sides of the above equation
It will be $2gh = 64g{e^2}$
$ \Rightarrow h = 32{e^2}$
Put \[e = 0.5\;\] (given)
It implies that, $h = 8{\text{ }}m$
Hence, option C is the correct answer i.e., ${\text{8 }}m$.
Note: When two bodies collide, the coefficient of restitution is the ratio of the two objects' final relative velocity to their starting relative velocity following contact. The coefficient of restitution of a bouncing ball is found by taking the square root of the ratio of one bounce to the height of the following bounce.
Formula used:
If the ball is falling from the height $H$, then its velocity is;
$v = \sqrt {2gH} $
Relation between velocity and coefficient of restitution –
${v_f} = ev$
Complete step by step solution:
Given that,
Steel ball dropped from a height \[32{\text{ }}m\].
Let, the height be $H = 32{\text{ }}m$
Now, the ball reaches the plate and bounces from the steel plate. Let the height to which the steel ball rebounds be $h$. As we know, If the ball is falling from the height $H$, then its velocity is
$v = \sqrt {2gH} $
Put $H = 32{\text{ }}m$
Velocity of steel ball when it strikes the plate will be $v = \sqrt {64g} $
After striking the plate, the steel ball rebounds with the velocity ${v_f}$.
Where, ${v_f} = ev$ ----------(1)
Here, $e = $ coefficient of restitution and $v = $ velocity of striking the plate
Also, final velocity will be ${v_f} = \sqrt {2gh} $ ---------(2)
From equation (1) and (2), we get
$\sqrt {2gh} = ev$
It can be written as $\sqrt {2gh} = e\sqrt {64g} $
Squaring both the sides of the above equation
It will be $2gh = 64g{e^2}$
$ \Rightarrow h = 32{e^2}$
Put \[e = 0.5\;\] (given)
It implies that, $h = 8{\text{ }}m$
Hence, option C is the correct answer i.e., ${\text{8 }}m$.
Note: When two bodies collide, the coefficient of restitution is the ratio of the two objects' final relative velocity to their starting relative velocity following contact. The coefficient of restitution of a bouncing ball is found by taking the square root of the ratio of one bounce to the height of the following bounce.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement
