Question #a8f56

1 Answer
Joaquín M.
May 12, 2016

10.05m

Explanation:

When the pen drops, the distance to the the ground is 10m

h=1m/s*10s = 10 m

and the pen has an initial speed of 1m/s, equal to the air ballon, direction towards up. Thus, during a short period of time the pen is going up until the speed is 0 for the action of gravity. So, to solve the problem it is necessary to calculate the distance pen went up from the moment the pilot dropped it.

The acceleration of the pen is a_y= -g where g is the acceleration of gravity and by integration

v_y = a_y*t+v_(yo) = -g*t+v_(yo)
y=-g*t^2/2+v_(yo)*t+y_0

The pen reaches the maximum high when v_y = 0, so the time it lasts after the pilot drops it is t=0.1s. Therefore, the distance that pen drops is

y= -10m/s^2*0.1^2*s^2+1m/s*0.1s+10m = 10.05m

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