A body is thrown up with a of 40m/s .how long does it take to reach the highest point ?
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Initial velocity (u) be 40m/s
final velocity (v) = 0 (as after reaching the highest point it's velocity would reduce to 0)
Acceleration = - acceleration due to gravity = -9.8m/s^2
By using 1rd eq. of motion
v = u+at
0 = 40 +(-9.8)(t)
-40 = -9.8t
t = -40/-9.8
= 4.081 s
Hence, it would take 4.081 s for the body to reach it's highest point.
final velocity (v) = 0 (as after reaching the highest point it's velocity would reduce to 0)
Acceleration = - acceleration due to gravity = -9.8m/s^2
By using 1rd eq. of motion
v = u+at
0 = 40 +(-9.8)(t)
-40 = -9.8t
t = -40/-9.8
= 4.081 s
Hence, it would take 4.081 s for the body to reach it's highest point.
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Answer:
Let us take g= 10N/m^2
distance = u^2/2g
= 40×40÷2×10
= 1600÷20
= 80
Therefore, It takes 80m to reach highest point
Time taken. = 2u÷g
= 2×40÷10
= 80÷10
= 8sec
The total time taken = 8sec
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