A body is projected vertically upward with speed 40 m/s . the distance travelled by body in the last second of upward journey is what ( g= 9.8 m/s and neglect eff8 of air resistance)

A body is projected vertically upward with speed 40 m/s . the distance

1.	A body is projected vertically upward with speed 40 m/s . the distance travelled by body in the last second of upward journey is what ( g= 9.8 m/s and neglect eff8 of air resistance)
Answer» HY.. Friend... Total distance travelled by body = H Distance travelled by body in last SECOND = d H = u^2 / (2g) = 40^2 / (2 * 9.8) = 1600 / 19.6 m = 81.63 m Time taken by body to REACH height H is sqrt(2H/g) = sqrt(2 * 81.63 / 9.8) = 4.08 seconds Use equation of motion..., S = ut + 0.5at^2 H - d = 40(t - 1) - 0.5g(t - 1)^2 81.63 - d = 40 (4.08 - 1) - 0.59.8(4.08 - 1)^2 81.63 - d = 76.71 d = 81.63 - 76.71 d = 4.9 m Thankyou

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A body is projected vertically upward with speed 40 m/s . the distance travelled by body in the last second of upward journey is what ( g= 9.8 m/s and neglect eff8 of air resistance)

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