12. An object is released from a height.
(a) Find its speed at (1) t = 1s, (2) t = 2s, (3) t = 3s.
(b) Find the distance traveled at (1) t = 1s, (2) t = 2s (3) t = 3s
As an object falls, its speed increases because it’s being pulled on by gravity. The acceleration of gravity near the earth is g = -9.81 m/s^2. To find out something’s speed (or velocity) after a certain amount of time, you just multiply the acceleration of gravity by the amount of time since it was let go of. So you get: velocity = -9.81 m/s^2 * time, or V = gt. The negative sign just means that the object is moving downwards. If it were positive, then it would be moving up. For speed rather than velocity, you just drop the negative sign.
(a) 1.speed @ t=1s
v = gt
v = -9.81 m/s^2 * 1 s
v = -9.81 m/s (moving downwards)
2. speed @ t=2s
v = gt
v = -9.81 m/s^2 * 2 s
v = -19.62 m/s (moving downwards)
3. speed @ t=3s
v = gt
v = -9.81 m/s^2 * 3 s
v = -29.43 m/s (moving downwards)
(2) 1. distance @ t=1s
h = 1/2 * gt^2
h = 0.5 * -9.81 m/s^2 * 1s
h = 4.903325 m
2. distance @ t=1s
h = 1/2 * gt^2
h = 0.5 * -9.81 m/s^2 * 2s
h = 19.6133 m
3. distance @ t=3s
h = 1/2 * gt^2
h = 0.5 * -9.81 m/s^2 * 3s
h = 44.129925 m
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