Question

A body is thrown vertically up with some velocity (at t = 0) crosses a fixed point P in its path at t = 4s and again at t = 6 s. The velocity of projection of the body will be ​

Answer 1

Since it takes (6-4) second in passing same point

So it takes 1 second for moving from that point to maximum height.

So time of ascent is (4+1) = 5second

V = u +at

0= u -gt

u= gt

u = 5g = 50m/s

Answer 2

Answer : 49 m/s

Given    : Vertically upward  implies θ =90

              thrown body passes fixed point  at t= 4s and t=6s

To find   : Velocity of projection

Explanation :

An object that is in flight after being thrown or projected is called a projectile.The path (trajectory) of a projectile is a parabole.

Displacement (h) of the projectile after a time t can be found using the equation

        h = u sinθ t − 0.5 $gt^{2}$

Where u is the velocity of projection θ is the angle of projection t is the time taken to  reach the point and g is the acceleration due to gravity.

Here displacement h is same for both t=4s and t=6s

Therefore

4u sin90 - 0.5*16*g = 6u sin90 - 0.5*36*g

              but sin90 = 1 substituting we get

                 4u - 8*g = 6u - 18*g

                        2u = 10*g

                          u = 5*g

                             = 5*9.8

                             = 49 m/s

Velocity of projection of body will be 49 m/s.

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