a body is projected with a speed of 40m/s vertically up from the ground. what is the maximum height reached by the body?what is hte entire time of motion? what is the velocity at 5 seconds after the projection?
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Maximum height
H = u^2 / (2g)
= 40^2 / (2 * 10)
= 80 m
Time of motion
T = sqrt(2H/g)
= sqrt(2 * 80 / 10)
= 4 seconds
Velocity at 5s after the projection
v = u + at
v = 40 + (-10 * 5)
v = -10 m/s
Velocity at 5s after the projection is 10 m/s vertically downwards, hence -ve sign.
H = u^2 / (2g)
= 40^2 / (2 * 10)
= 80 m
Time of motion
T = sqrt(2H/g)
= sqrt(2 * 80 / 10)
= 4 seconds
Velocity at 5s after the projection
v = u + at
v = 40 + (-10 * 5)
v = -10 m/s
Velocity at 5s after the projection is 10 m/s vertically downwards, hence -ve sign.
Answered by
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Answer:Maximum height
H = u^2 / (2g)
= 40^2 / (2 * 10)
= 80 m
Time of motion
T = sqrt(2H/g)
= sqrt(2 * 80 / 10)
= 4 seconds
Explanation: please mark as a brain list
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