In an imaginary atmosphere, the air exerts a small force F on any particle in the direction of the particle's motion. A particle of mass m projected upwards takes a time t1 in reaching the maximum height and t2 in the return journey to the original point. Then: A. t1t2 C. t1=t2

The force exerted by the air on the particle aids the particle’s speed. When a particle goes upwards, we know that the direction of gravitational force is opposing the direction of motion. The particle is attracted by the gravitational attraction, which slows down the particle’s upward velocity. Both gravitational force and air force operate in the direction of motion as a particle goes downhill, implying that they promote the motion. As a result, we may state that the particle’s downhill speed is larger than the particle’s upward speed.


We also know that a particle’s travelling time is given by the ratio of the distance traveled by it and its speed. Mathematically, we can


t =d/s

Here t is the traveling time, d is the distance traveled, and s is the particle’s speed.

Write the expression for the traveling time of the particle for upward motion.

t1 = d/s1 ……(1)

Write the expression for the traveling time of the given particle for downward motion.

t2 = d/s2 ……(2)

On dividing equation (1) and equation (2), we can write:

t2/t1 = s1/s2

We already know that s2>s1

so t1>t2 ..

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