A body is projected vertically upward with speed 40m/s. The distance travelled by body in the ladt second of upward journey is?
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Answered by
76
Distance travelled by body in last second of “upward” journey = Distance travelled by body in first second of “downward” journey.
Distance travelled by body in first second of “downward” journey
S = 0.5at^2
= 0.5 * 9.8 * 1^2
= 4.9 m
∴ Distance travelled by body in last second of upward journey is 4.9 m (or 5 m if g = 10 m/s^2)
Distance travelled by body in first second of “downward” journey
S = 0.5at^2
= 0.5 * 9.8 * 1^2
= 4.9 m
∴ Distance travelled by body in last second of upward journey is 4.9 m (or 5 m if g = 10 m/s^2)
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3
The distance travelled in last seco d will be equal to the distance travellad in first second
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