physics
posted by mona
A body is projected vertically upwards with a speed of 40 m/s.Find the distance travelled by body in last second of upward journey?(g=9.8m/s2 neglect effect of air resistance)

Henry
V = Vo + g*Tr = 0 at max. ht.
Tr = Vo/g = 40/9.8 = 4.08 s. = Rise
time.
T = 4.081 = 3.08 s.
h1 = Vo*t + 0.5g*t^2
h1 = 40*3.08  4.9*3.08^2 = 76.7 m.
h2 = 40*4.08  4.9*4.08^2 = 81.6 m.
h2h1 = 81.676.7 = 4.93 m. = Distance traveled in the last second.
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A body is projected vertically upward with a speed of 40 m/s. Find the distance travelled by the body in the last sec of the upward journey taking g = 9.8m/s