Assertion : A tennis ball bounces higher on h | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the tennis ball bounce with an initial velocity $u$. It will rise to a height $h$ where its final velocity becomes $0$. Using the equation of

Vedclass · Accepted Answer

If the $Assertion$ is correct but $Reason$ is incorrect.

Vedclass · Answer

(C) Let the tennis ball bounce with an initial velocity $u$. It will rise to a height $h$ where its final velocity becomes $0$. Using the equation of motion $v^2 - u^2 = 2as$,we get:$0^2 - u^2 = 2(-g')h$,where $g'$ is the acceleration due to gravity on the hill.Therefore,$h = \frac{u^2}{2g'}$.Since the acceleration due to gravity on the hill $(g')$ is less than that on the surface of the earth $(g)$ due to the increase in altitude,the height $h$ will be greater on the hill.Thus,the $Assertion$ is correct,but the $Reason$ is incorrect because $g'$ is actually less than $g$.

$Assertion$ : $A$ tennis ball bounces higher on hills than in plains.
$Reason$ : Acceleration due to gravity on the hill is greater than that on the surface of earth.

Similar Questions

What is the value of acceleration due to gravity $(g)$ at the centre of the Earth? What will be the variation of $g$ below and above the surface of the Earth?

Two planets $A$ and $B$ have the same mass $M$ and radius $R$. The variation of density $\rho$ of the planets with distance $r$ from the center is shown in the following diagrams. The ratio of acceleration due to gravity at the surface of the planets $A$ and $B$ will be:

The radius of the Earth is $6400 \,km$ and the acceleration due to gravity is $g=10 \,ms^{-2}$. For the weight of a body of mass $5 \,kg$ to be zero at the equator,the rotational angular velocity of the Earth must be (in $rad/s$):

The value of acceleration due to gravity at a height of $4 R_E$ from the surface of the Earth is (where $R_E$ is the radius of the Earth and acceleration due to gravity on the surface of the Earth $g = 10 \,ms^{-2}$): (in $\,ms^{-2}$)

At what height above the surface of the Earth does the value of $g$ decrease by $2 \%$? [Radius of the Earth is $6400 \, km$]

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$Assertion$ : $A$ tennis ball bounces higher on hills than in plains.$Reason$ : Acceleration due to gravity on the hill is greater than that on the surface of earth.