Question

a body is projected vertically up at t=0 with velocity 50m/s. another body is projected from same point with same velocity after 3second, both body will meet at t=___​

Answer 1

Answer:

19.80 seconds

Explanation:

Given:

The velocity, u = 50 m/s

Time difference for the second body = 3 seconds

let 't' be the time taken by the first object after which the first object meets the second object.

therefore, the time taken by the second object will be (t - 3) seconds.

Now, for both the objects to meet, the distance covered by them should be equal.

from the Newton's equation of motion, we have

$s=ut+\frac{1}{2}(-g)t^2$

here,

g is the acceleration due to the gravity and the negative sign depicts that the acceleration due to gravity is in opposite direction to the movement of the object.

Thus,

$s=50\times t+\frac{1}{2}(-9.8)\times t^2$

and for the second object

$s=50\times (t-3)+\frac{1}{2}(-9.8)\times (t-3)^2$

equation both the equations , we get

$50\times t+\frac{1}{2}(-g)\times t^2=s=50\times (t-3)+\frac{1}{2}(-g)\times (t-3)^2$

or

50t - 4.9t² = 50t - 150 - 4.9t² - 44.1 + 9.8t

or

9.8t - 194.1 = 0

or

t = 19.80 seconds