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A hot air balloon competition requires a balloonist to drop a ribbon onto a target on the ground. Initially the hot air balloon is 50 meters above the ground and 100 meters from the target. The wind is blowing the balloon at v = 15 meters/sec on a course to travel directly over the target. The ribbon is heavy enough that any effects of the air slowing the vertical velocity of the ribbon are negligible. How long should the balloonist wait to drop the ribbon so that it will hit the target?

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Answer:

The answer is 3.48 seconds

Explanation:

The kinematic equation

y= y0+V0*t+1/2*a*(t*t)

-50=0+(0)t+1/2(-9.8)*(t*t)

t=3.194 seconds

During ribbons ball,

x=x0+ Vt+1/2*a*(t*t)

x= 0+(15)*(3.194)+1/2*(0)* (3.194*3.194)

x= 47.9157m

So, distance (D) = 100-47.9157= 52.084m

52.084m=0+15(t)+1/2*(0)(t*t)

t=52.084/15=3.472286= 3.48seconds

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