Question:

A body dropped from a height h with initial velocity zero, strikes the ground with a velocity 3 m/s. Another body of same mass dropped from the same height h with an initial velocity of 4 m/s. The final velocity of second mass, with which it strikes the ground is

Updated On: Jul 29, 2024
  • 5 m/s
  • 12 m/s
  • 3 m/s
  • 4 m/s
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The Correct Option is A

Solution and Explanation

Initial velocity of first body $(u_1) = 0 ;$
Final velocity $(v_1) = 3\, m/s$ and initial velocity of
second body $(u_2) = 4\, m/s$
height $(h)=\frac{v_1^2}{2g}=\frac{(3)^2}{2 \times9.8}=0.46\, m.$
Therefore velocity of the second body,
$V_2=\sqrt{u_1^2+2gh}=\sqrt{(4)^2+2 \times9.8 \times 0.46}{}=5\, m/s.$
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Concepts Used:

Motion in a straight line

The motion in a straight line is an object changes its position with respect to its surroundings with time, then it is called in motion. It is a change in the position of an object over time. It is nothing but linear motion. 

Types of Linear Motion:

Linear motion is also known as the Rectilinear Motion which are of two types:

  1. Uniform linear motion with constant velocity or zero acceleration: If a body travels in a straight line by covering an equal amount of distance in an equal interval of time then it is said to have uniform motion.
  2. Non-Uniform linear motion with variable velocity or non-zero acceleration: Not like the uniform acceleration, the body is said to have a non-uniform motion when the velocity of a body changes by unequal amounts in equal intervals of time. The rate of change of its velocity changes at different points of time during its movement.