A body is projected vertically upward. Show the following.
(i) The distance covered by it in the first $n$ seconds of ascent is equal to the distance covered by it in the last $n$ seconds of its descent.
(ii) The distance covered by it in the last $n$ seconds of ascent is equal to the distance covered by it in the first $n$ seconds of its descent.

Open in App

Solution

Find the velocity $v_{1}$ of a body projected upwards after $t = (t_{1} - n)$ seconds
$t_{1} = t i m e o f a s c e n t = \frac{u_{1}}{g}$ using equation
$v_{1} = u_{1} - g t$

Suggest Corrections

Similar questions

Q. A body is projected up with a velocity

96 m / s

. The ratio of distance covered by the body in last two seconds of its motion and last two seconds of its ascent.

Q. A stone falls freely such that the distance covered by it in the last second of its motion is equal to the distance covered by it in the first

5

seconds. It remained in air for:

A ball is dropped from height h.The total distance covered by the body in last second of its motion is equal to distance covered by it in the first three seconds ,what is the vslue of h.

Q. The distances covered by a freely falling body in its first, second, third,......

n^{t h}

seconds of its motion

Q. A stone falls freely from rest and the total distance covered by it in the last second of its motion equals the distance covered by it on the first three seconds of its motion. The stone remains in the air for

Join BYJU'S Learning Program

Find the velocity v1 of a body projected upwards after t=(t1−n) secondst1=time of ascent=u1g using equationv1=u1−gt

Find the velocity $v_{1}$ of a body projected upwards after $t = (t_{1} - n)$ seconds
$t_{1} = t i m e o f a s c e n t = \frac{u_{1}}{g}$ using equation
$v_{1} = u_{1} - g t$