Astudent launches a small rocket which starts from rest and from ground level (a height of h = 0). at a height of h = 1.95 km the rocket has reached a velocity of vf = 389 m/s. at that height the rocket runs out of fuel and therefore there is no longer any thrust propelling the rocket. you may assume that the positive direction is upwards in this problem.

during the upward motion of rocket till its fuel is not over the acceleration is given by

now the time taken by the rocket to reach that point is given as

After fuel is over the maximum height that the rocket will reach is given by

so the total height of rocket from ground is given by

now the time taken by the rocket to reach back on ground is

total time of motion of the rocket will be

T = 10 + 84.11 = 94.11 s

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i just did this and got a 100.

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be