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?
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
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(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
To find the time it takes for the camera to hit the ground, we can use the equation of motion for objects in freefall:
h = (1/2) * g * t^2
Where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
In this case, the height is 15.0 m, so we can rearrange the equation to solve for t:
t = sqrt((2 * h) / g)
Plugging in the values:
t = sqrt((2 * 15.0) / 9.8)
t ≈ sqrt(3.06)
t ≈ 1.75 seconds
So it takes approximately 1.75 seconds for the camera to hit the ground.
To find the speed of the camera just before it hits the ground, we can use another equation of motion:
v = g * t
Where v is the velocity (speed) and t is the time.
Plugging in the values:
v = 9.8 * 1.75
v ≈ 17.15 m/s
So the speed of the camera just before it hits the ground is approximately 17.15 m/s.
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