Answer :

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/s^{2} (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/su(initial velocity) = 29.4 m/s

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

^{2}

using third equation of motion , we get

v^{2} = u^{2} + 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 secu(initial velocity)=0 m/s

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

^{2}

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

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