# A physics problem..help?

Question:An elevator without a ceiling is ascending with a constant speed of 10 m/s. A boy on the elevator throws a ball directly upward, from a height of 2.0 m above the elevator floor, just as the elevator floor is 31 m above the ground. The initial speed of the ball with respect to the elevator is 17 m/s.
1.What is the maximum height attained by the ball?
2.How long does it take for the ball to return to the elevator floor?

show work if possible..

The ball is projected from a point (31 + 2)m above the ground, with an initial velocity relative to the ground of (10 + 17)m/sec.

1.
If h is the maximum height from where the ball is released, then:
0^2 = (10 + 17)^2 - 2gh
2gh = 27^2
h = 27^2 / 2g
Using g = 9.81 m/sec^2:
h = 27^2 / (2*9.81)
= 37.16m.
The ball was 33m above ground when released, and so the maximum height above ground is 33 + 37.16m = 70.16m.

2.
It's unclear where the time measurement is to start. I am assuming it is from the moment the ball starts descending.

The time t from the moment the ball is thrown up at 27m/sec until it starts descending is given by:
0 = 27 - gt
t = 27 / g = 27/9.81 = 2.75sec.
In that time, the lift floor has ascended 10*2.75 = 27.5m from its initial height of 31m.
It is therefore 58.5m above ground.

At time t after the ball starts descending, its height s above ground is:
s = 70.16 - gt^2 / 2

The height s1 of the lift floor above ground at the same time is:
s1 = 58.5 + 10t

The required time t is therefore when:
s = s1
70.16 - gt^2 / 2 = 58.5 + 10t
gt^2 / 2 + 10t - 11.66 = 0
4.905t^2 + 10t - 11.66 = 0
t = ( -10 +/- sqrt(100 + 4 * 11.66 * 4.905) ) / 2
= ( - 10 +/- 18.13 ) / 2.

t = 8.13 / 2 = 4.07 sec.

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