# Need a help with section C of this question guys.Watch

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#1
An object is released, 50m above the ground from a hot air balloon, which is descending vertically at a speed of 4.0 ms^. Calculate :
a) the velocity of the object at the ground(i've done this part and the answer for it is 32ms^)
b)the duration descent of the object (i've done this part as well and the answer for it is 2.8s)
C) the height of the ballon above the ground when the object hits the ground.
(please explain to me part c as clear as possible. i know the answer for it but i can't understand the question)
tnx guys .
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4 years ago
#2
well you know how long it took for the object to reach the ground... during that time the balloon has moved the balloon at a constant speed of 4m/s downwards.

so use D,S,T to find the distance the balloon has moved.... then subtract that from the original height above the ground.
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#3
sorry just didn't get what u mean by D,S,T? if you mean distance , time and speed then where do u get the value of them from?
i appreciate your help tho .
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4 years ago
#4
(Original post by Alen.m)
sorry just didn't get what u mean by D,S,T? if you mean distance , time and speed then where do u get the value of them from?
i appreciate your help tho .
you have worked out T

you are told S

you can find D
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4 years ago
#5
Let's look at the situation: the hot air balloon is 50m in the air. It falls at a rate of 4m/s for some time T. Could you write an expression for how high it would be after that time?
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#6
i understand that i need to multiply 4s by 2.8ms^ in order to find out the distance of ballon from an object which is 11.2m but why would you take that away from 50 to get the answer?
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4 years ago
#7
(Original post by Alen.m)
i understand that i need to multiply 4s by 2.8ms^ in order to find out the distance of ballon from an object which is 11.2m but why would you take that away from 50 to get the answer?
Because 11.2m is relative to the balloon's original position, it has fallen by that much. Unfortunately, the question actually asks for it's current height relative to the ground which is not the same thing.
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#8
thanks for your time mate .
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