Question

Q.2. A Man is standing at the top of the 60 m high tower. He throws the ball vertically upwards with the velocity of 20 m/s. After what time will the ball pass him goes downwards? How long after its release will the ball reach the ground?

Answer 1

Dear Student

Height of the tower, h = 60 m
Initial velocity, u = 20 m/s
Final velocity, v = 0

We know,
v = u - gt
0 = 20 - 10t
So, 10t = 20
t = 2s

Now, time of ascent = time of descent
so, the ball passes the man after 4s going downwards
Therefore, using -
s = ut + $\frac{1}{2}$ gt²
60 = 20t + $\frac{1}{2}$ 10t^​2
dividing by 10, from both the ends -
t² + 4t - 12 = 0
t² + 6t - 2t- 12 = 0
t (t + 6) - 2 (t + 6) = 0
So, (t + 6) (t - 2) = 0

So, t = -6, which cannot be possible
Therefore, t will be equal to 2

Thus, the time taken by the ball to reach the ground after release = (4 + 2) s= 6 sec


Regards