A ball thrown up vertically returns to the thrower after 6 s, find:

(a) The velocity with which it was thrown up,

(b) The maximum height it reaches, and

(c) Its position after 4 s.

Since the ball thrown up vertically returns to the thrower in 6 seconds, this means that the ball will take half of this time.

= 3 s

(a) Given,

Final velocity, v = 0

Initial velocity, u =?

Acceleration due to gravity, g = -9.8 m/s² (As the ball goes up)

Time taken, t = 3 s

We know that,

v = u + g × t

0 = u + (-9.8) × 3

0 = u – 29.4

u = 29.4 m/s

(b) Now, calculation of maximum height:

We know that, at maximum height velocity of object will be Zero 

v(final velocity at max height)=0 m/s

u(initial velocity)  = 29.4 m/s

g(acceleration due to gravity)=10 m/s²

using third equation of motion , we get

v² = u² + 2gh

(0)² = (29.4)² + 2 × (-9.8) × h

0 = 864.36 – 19.6 h

19.6 h = 864.36

h =

h = 44.1 m

(c) Finally, we have to calculate the position of ball after 4 seconds:

We know that, the ball will at max height when t=3 sec at t = 4 sec the ball has been falling for 1 sec from the max height  

hence, t= 1 sec

u(initial velocity)=0 m/s

a(acceleration due to gravity) = 9.8 m/s²

h = 0×1+ ×9.8 × (1)² (Because t = 1 s)

= 0 + 4.9 ×1

= 4.9 m
Hence, ball has been fallen 4.9 m from the max height

The height from the ground= maximum height - height object has fallen in 1 sec= 44.1- 4.9= 39.2 m above the ground