
A ball P is dropped vertically and another ball Q is thrown horizontally with the same velocities from the same height and at the same time. If air resistance is neglected, then
A. Ball P reaches the ground first
B. Ball Q reaches the ground first
C. Both reach the ground at the same time
D. The respective masses of the two balls will decide the time
Answer
218.4k+ views
Hint:When a ball is dropped vertically and another ball is thrown horizontally with the same height, same velocities and same time, then initial velocity for both cases will be zero. Also for each case acceleration due to gravity, $g$ acts towards the ground. So, we can find out the time of reaching the ground by using the third equation of motion.
Formula used:
From the third equation of motion, we can write the formula:
Displacement,$s=ut+\dfrac{1}{2}g{{t}^{2}}$
Here $u=$Initial velocity
$t$ and $g$ is the time and acceleration due to gravity.
Complete step by step solution:
As both the balls were initially in rest position. Hence the value of $u$ is zero.
Thus, by putting this value into the equation the third of motion we can get,
$s=0\times t+\dfrac{1}{2}g{{t}^{2}}=\dfrac{1}{2}g{{t}^{2}}$
$\Rightarrow {{t}^{2}}=\dfrac{2s}{g}$
Equation for time required by ball to reach the ground is given by,
$t=\sqrt{\dfrac{2s}{g}}$
Here the displacement of the ball is the height from which it is thrown. Hence, $s = h$
So, $t=\sqrt{\dfrac{2h}{g}}$
The above equation signifies that the time taken by ball to reach the ground depends only on the height h from the ball is thrown irrespective of the way it is thrown horizontally and vertically. $g$ is a constant. As the height from which the ball is thrown is the same so the two balls will reach the ground at the same time.
Therefore, the correct option is C.
Note: When a ball is thrown vertically its kinetic energy keeps on decreasing and at maximum height velocity is zero. But potential energy keeps on increasing and at maximum height only potential energy will exist. Also when the same ball is dropped vertically the potential energy keeps on decreasing and kinetic energy increases gradually.
Formula used:
From the third equation of motion, we can write the formula:
Displacement,$s=ut+\dfrac{1}{2}g{{t}^{2}}$
Here $u=$Initial velocity
$t$ and $g$ is the time and acceleration due to gravity.
Complete step by step solution:
As both the balls were initially in rest position. Hence the value of $u$ is zero.
Thus, by putting this value into the equation the third of motion we can get,
$s=0\times t+\dfrac{1}{2}g{{t}^{2}}=\dfrac{1}{2}g{{t}^{2}}$
$\Rightarrow {{t}^{2}}=\dfrac{2s}{g}$
Equation for time required by ball to reach the ground is given by,
$t=\sqrt{\dfrac{2s}{g}}$
Here the displacement of the ball is the height from which it is thrown. Hence, $s = h$
So, $t=\sqrt{\dfrac{2h}{g}}$
The above equation signifies that the time taken by ball to reach the ground depends only on the height h from the ball is thrown irrespective of the way it is thrown horizontally and vertically. $g$ is a constant. As the height from which the ball is thrown is the same so the two balls will reach the ground at the same time.
Therefore, the correct option is C.
Note: When a ball is thrown vertically its kinetic energy keeps on decreasing and at maximum height velocity is zero. But potential energy keeps on increasing and at maximum height only potential energy will exist. Also when the same ball is dropped vertically the potential energy keeps on decreasing and kinetic energy increases gradually.
Recently Updated Pages
Two discs which are rotating about their respective class 11 physics JEE_Main

A ladder rests against a frictionless vertical wall class 11 physics JEE_Main

Two simple pendulums of lengths 1 m and 16 m respectively class 11 physics JEE_Main

The slopes of isothermal and adiabatic curves are related class 11 physics JEE_Main

A trolly falling freely on an inclined plane as shown class 11 physics JEE_Main

The masses M1 and M2M2 M1 are released from rest Using class 11 physics JEE_Main

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

Understanding Atomic Structure for Beginners

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

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

NCERT Solutions For Class 11 Physics Chapter 8 Mechanical Properties Of Solids

Motion in a Straight Line Class 11 Physics Chapter 2 CBSE Notes - 2025-26

NCERT Solutions for Class 11 Physics Chapter 7 Gravitation 2025-26

How to Convert a Galvanometer into an Ammeter or Voltmeter

