a body is thrown upward with a velocity of 20 m per second how much time will it take to return to its original position
Sunandit:
4 sec.
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Initial velocity U = 20 m/s upward.
Final velocity V = -20m/s downward.
Acceleration a = - 9.8 m/s^2 Since this acts downward .
Using in V = U + a T
T = (V- U) / a. Note that the definition of acceleration is a = (v-u)/t .
T = (-20 - 20) / - 9.8 = (40) / 9.8 = 4.08 second.
Since the object returns to the initial position the displacement is zero.
Or use V^2 - U^2 = 2a s. Since both v^ 2 and U^2 are the same, S = 0.
To find the distance traveled, we calculate twice the height through which it goes up.
In this case at the highest position V= 0.
S = V^2 / 2g = 200 / 9.8 = 20.4m
Total distance traveled = 2 x 20.4= 40.8 m.
Final velocity V = -20m/s downward.
Acceleration a = - 9.8 m/s^2 Since this acts downward .
Using in V = U + a T
T = (V- U) / a. Note that the definition of acceleration is a = (v-u)/t .
T = (-20 - 20) / - 9.8 = (40) / 9.8 = 4.08 second.
Since the object returns to the initial position the displacement is zero.
Or use V^2 - U^2 = 2a s. Since both v^ 2 and U^2 are the same, S = 0.
To find the distance traveled, we calculate twice the height through which it goes up.
In this case at the highest position V= 0.
S = V^2 / 2g = 200 / 9.8 = 20.4m
Total distance traveled = 2 x 20.4= 40.8 m.
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