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Boxes of toys are dropped to children from a helicopter travelling at a speed of 80.0 m/sin a direction of 50.0° above the horizontal. The helicopter is at a height of 200 m fromthe playground. Calculate the distance at which the pilot should release the boxes so thatthe boxes land at the point where the children are standing(a) 187 m(b) 138 m(c) 781 m(d) 168 m

Question
Boxes of toys are dropped to children from a helicopter travelling at a speed of 80.0 m/s
in a direction of 50.0° above the horizontal. The helicopter is at a height of 200 m from
the playground. Calculate the distance at which the pilot should release the boxes so that
the boxes land at the point where the children are standing
(a) 187 m
(b) 138 m
(c) 781 m
(d) 168 m
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Boxes of toys are dropped to children from a helicopter travelling at a speed of 80.0 m/s in a direction of 50.0° above the horizontal. The helicopter is at a height of 200 m from the playground. Calculate the distance at which the pilot should release the boxes so that the boxes land at the point where the children are standing (a) 187 m (b) 138 m (c) 781 m (d) 168 m

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Step 1

Consider the distance at which the pilot should release the boxes be x, the time taken by the boxes to reach the children be t, the height or distance of the helicopter from the playground be h, and the speed of the helicopter be u.

 

The horizontal and vertical components of the speed of the helicopter are,

ug ucos50°=(80 m/s)cos 50
u, usin 50° (80 m/s)sin 50°
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ug ucos50°=(80 m/s)cos 50 u, usin 50° (80 m/s)sin 50°

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Step 2

Write the equation of motion, and solve for the time taken by the boxes to reach the children.

1
h ut
2
(9.81 m/s )r
(-200 m)(80 m/s) sin 50°)t
2
4.9052-61.28t - 200 = 0
(-61.28) - 4(4.905)(-200)
2(4.905)
-(-61.28).
t =
t-(15.179 s), 2.68 s)
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1 h ut 2 (9.81 m/s )r (-200 m)(80 m/s) sin 50°)t 2 4.9052-61.28t - 200 = 0 (-61.28) - 4(4.905)(-200) 2(4.905) -(-61.28). t = t-(15.179 s), 2.68 s)

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Step 3

Take the positive value of the time, and write the expression for the relation between the speed and ti...

Hn
t
(80 m/s) cos 50°(15.179 s)
x
x = 781 m
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Hn t (80 m/s) cos 50°(15.179 s) x x = 781 m

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