A ball is dropped from a balloon going up at a speed of 7m/s. If the balloon was at a height of 60m at the time of dropping the ball, how long will the ball take in reaching the ground?
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Answers
The ball will take:
t = v/9.8 = 7/9.8 = 0.7 seconds to reach its apex if thrown upwards at 7 ms-1.
During this time it will cover:
s = 1/2 a x t ^2
s = 4.9 x 0.5 = 2.45m
So after 0.7 seconds it has zero velocity and then has to travel through 60m + 2.45m = 62.45m.
It will need:
s =1/2 a x t^2
62.45 = 1/2 x 9.8 x t^2
t = 3.57 seconds to do this.
Adding 3.57 to 0.7 seconds = 4.27 answer.
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Explanation:
Let’s measure distances down towards the ground from the point where the ball is dropped.
We know: the distance (s=60); the initial velocity (u=−7 where the negative value reflects the fact that the ball is moving upwards when dropped); the acceleration (a=g≈9.81); and we want to find: the time (t). The relevant equation of motion we need, relating these quantities, is:
s=ut+12at2
∴60≈−7t+4.91t2 i.e. 4.91t2−7t−60≈0
Recall that we can solve a general quadratic equation ax2+bx+c=0 using the formula x=12a(−b±b2−4ac−−−−−−−√). We can apply this formula with a=4.91, b=−7, c=−60, as follows:
t≈19.81(7±49+1177.2−−−−−−−−−√)≈0.7136±3.5695=4.2831 or −2.8559
The negative solution will represent a time at which the ball could have left the ground, heading upwards, to reach the balloon at t=0, so we can ignore that solution to the equation. The solution we require is t≈4.2831.
So the ball will take about 4.28 seconds to reach the ground.