(C) When an object is dropped from a height $h$ under gravity,its motion is governed by the equation of motion $h = ut + \frac{1}{2}gt^2$. Since the i

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

Both stones reach the ground simultaneously.

(C) When an object is dropped from a height $h$ under gravity,its motion is governed by the equation of motion $h = ut + \frac{1}{2}gt^2$. Since the initial velocity $u = 0$,the equation simplifies to $h = \frac{1}{2}gt^2$. Solving for time $t$,we get $t = \sqrt{\frac{2h}{g}}$. As the time of flight $t$ depends only on the height $h$ and the acceleration due to gravity $g$,and is independent of the mass of the object,both stones will reach the ground simultaneously.

Class 11 Physics Motion in Straight Line Motion Under Gravity

Medium

Two stones of different masses are dropped simultaneously from the top of a building. Which of the following statements is true regarding their motion?

A
The smaller stone hits the ground earlier.
B
The larger stone hits the ground earlier.
C
Both stones reach the ground simultaneously.
D
Which stone reaches the ground earlier depends on the composition of the stone.

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

Motion in Straight Line

Subtopic

Motion Under Gravity

Two balls $A$ and $B$ are thrown with the same velocity $u$ from the top of a tower. Ball $A$ is thrown vertically upwards and ball $B$ is thrown vertically downwards. If $t_A = 6 \ s$ and $t_B = 2 \ s$,then the height of the tower is ......... $m$.

Medium

View Solution

$A$ ball is thrown vertically upwards with a velocity of $20 \; m s^{-1}$ from the top of a multistorey building. The height of the point from where the ball is thrown is $25.0 \; m$ from the ground. How long will it be in $seconds$ before the ball hits the ground? Take $g = 10 \; m s^{-2}$.

Medium

View Solution

An object is projected upwards from the foot of a tower. The object crosses the top of the tower twice with an interval of $8 \ s$ and the object reaches the foot again after $16 \ s$. The height of the tower is [ $g = 10 \ m/s^2$ ]. (in $m$)

EasyAP EAMCET 2024

View Solution

$A$ balloon rises from rest with a constant acceleration $g/8$. $A$ stone is released from it when it has risen to height $h$. The time taken by the stone to reach the ground is

Medium

View Solution

$A$ particle is projected from the ground in a vertically upward direction at $t = 0$,with an initial velocity of $48\, m/s$. The distance travelled by the particle in the $5^{th}$ second is......$m$ $(g = 10\, m/s^2)$.

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Two stones of different masses are dropped simultaneously from the top of a building. Which of the following statements is true regarding their motion?

Similar Questions

Two balls $A$ and $B$ are thrown with the same velocity $u$ from the top of a tower. Ball $A$ is thrown vertically upwards and ball $B$ is thrown vertically downwards. If $t_A = 6 \ s$ and $t_B = 2 \ s$,then the height of the tower is ......... $m$.

$A$ ball is thrown vertically upwards with a velocity of $20 \; m s^{-1}$ from the top of a multistorey building. The height of the point from where the ball is thrown is $25.0 \; m$ from the ground. How long will it be in $seconds$ before the ball hits the ground? Take $g = 10 \; m s^{-2}$.

An object is projected upwards from the foot of a tower. The object crosses the top of the tower twice with an interval of $8 \ s$ and the object reaches the foot again after $16 \ s$. The height of the tower is [ $g = 10 \ m/s^2$ ]. (in $m$)

$A$ balloon rises from rest with a constant acceleration $g/8$. $A$ stone is released from it when it has risen to height $h$. The time taken by the stone to reach the ground is

$A$ particle is projected from the ground in a vertically upward direction at $t = 0$,with an initial velocity of $48\, m/s$. The distance travelled by the particle in the $5^{th}$ second is......$m$ $(g = 10\, m/s^2)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module