Question

A body is projected up with a velocity $96m/s$ . The ratio of distance covered by the body in last two seconds of its motion and last two seconds of its ascent.

• A. $1:7$
• B. $7:9$
• C. $9:1$
• D. $3:5$

Answer 1

The correct option is A $1:7$

Velocity of a particle during the last second of its ascent will be $9.8m/s$

And the velocity of the same particle in the second before that will be $19.6m/s$

So the distance covered in both the seconds will be

$9.8+19.6m=29.4m$

Velocity of a particle during the last second of its descent will be $96m/s$

And the velocity of the same particle in the second before that will be $87.2m/s$

So the distance covered in both the seconds will be

$96+87.2m=183.2m$

The ratio of the distance covered during ascent and descent will be,

$=\frac{29.4}{183.2}$

$\approx \frac{1}{7}$