# Thread: Integration with respect to time...

1. ## Integration with respect to time...

Hey guys.

Could you please guide me through this? I don't get it at all.

A ball is dropped from a lookout 180 m high. At the same time, a stone is fired vertically upwards from the valley floor with speed V m/s. Take g = $\displaystyle 10m/s^2$. Find for what values of V a collision in the air will occur. Find, in terms of V, the time and the height when collision occurs, and prove that the collision speed is V m/s. Find the value of V for which they collide halfway up the cliff and the time taken.

2. Originally Posted by UltraGirl
Hey guys.

Could you please guide me through this? I don't get it at all.

A ball is dropped from a lookout 180 m high. At the same time, a stone is fired vertically upwards from the valley floor with speed V m/s. Take g = $\displaystyle 10m/s^2$. Find for what values of V a collision in the air will occur. Find, in terms of V, the time and the height when collision occurs, and prove that the collision speed is V m/s. Find the value of V for which they collide halfway up the cliff and the time taken.
if they are colliding halfway up a cliff, then both objects travel 90m, therefore, if the ball is dropped, it will take, where:

s = 1/2 at^2
90 = 0.5 x 10 x t^2
t = 4.24 seconds

then, using this benchmark and appreciating the time taken for the stone fired upwards to be 4.24 to reach 90m up, either the stone has gone higher, and is already on its way down, or it is merely near the peak of its flight path

from there, if the stone travels 90m, use this formula:

s = vt - 1/2 at^2
90 = 4.24v - 0.5 x 10 x 4.24^2
90 = 4.24v - 90
180 = 4.24v
v = 42.45 m/s

stone would be fired at 42.45m/s

3. Assuming they collide half way up the cliff is, of course, only the last part of the problem.

The dropped ball drops with acceleration $\displaystyle -10 m/s^2$, velocity, -10t m/s, and height $\displaystyle 180- 5 t^2$ m at time t seconds.

The ball fired upward also has acceleration $\displaystyle -10 m/s$ but with velocity V- 10t m/s and height $\displaystyle Vt- -5t^2$ at t seconds. In order to collide, they must have the same position at the same time: $\displaystyle Vt- 5t^2= 180- 5t^2$. The $\displaystyle -5t^2$ terms cancel leaving Vt= 180 or t= 180/V. Solve that for t, the time the balls collide. You can then use the formulas for the height at which the balls collide. Both time and height will depend upon V. Of course, the height must be between 0 and 180 m. That will give conditions on V such that they do collide. Set the height equal to 90 to get V such that the collide halfway up the cliff.

Finally, use the velocity equation to determine the velocity of each ball at the time of collision. The sum of the absolute values of the balls (one velocity will be positive, the other negative) will be the "collision speed".