
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
162.9k+ 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
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

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

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

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

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

Charging and Discharging of Capacitor

Other Pages
NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

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

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

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

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

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