Question

14.To test the durability of a new shock-proof camera, the company has the camera dropped from a height of 15.0 m. Assume air resistance is negligible. How long does it take for the camera to hit the ground? What is the speed of the camera just before it hits the ground?

Answer 1

T=sqrt(2 times height) *gravity
T= sqrt{(30)*9.8}
T= sqrt(3.06)
T= 1.7 seconds
Therefore it would take 1.7 seconds

Max height :
1.7 seconds * 9.8 m/s^2
=16.66 m or 17 meters approx
Therefore the max height reached is approx 17 m

Answer 2

(1/2) m v^2 = m g h
v = sqrt (2 g h) (not T)
v = 17.1 (not 1.7 )

average speed during fall = 17.1/2 = 8.57 m/s
time to fall = 15/8.57 = 1.75 seconds

Kim - it is dropped, not going up

Answer 3

a)d=vi + a (t)
12=0 +0.5 (-9.8) (t)^2
15=-4.9t^2
1.7s=t
b)vf=vi +a (t)
=0+ (-9.8)(1.7)
=16.66m/s approx 17m/s

Answer 4

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