A body is projected vertically upward with speed 40 m/s . The distance travelled by body in the last second of upward journey is .....
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Distance traveled by a body in last second of its upward journey= dist travelled by it in 1st second
u=40m/s. t=1s a=-10m/s2
Now
S=ut +1/2 at2
Putting values
S= 40×1 + 1/2×10×1.
S=45m
rashmipriyem3:
The Answer Given is 4.9 m ! I Dont Get It !
So it would be like
S= 0 + 1/2 × 9.8 × 1×1
Which is equal to 4.9
Thank You !
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