Question

A body is projected with a velocity of 40m/s .the distance travelled by body in the last second of the upward journey is

Answer 1

Hey
Here is your answer
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first of all we are given that
u=40m/s
a=10m/s
v=0m/s
now
v=u+at
0=40+(-10)t
t=4s
so total time for journey is 4seconds
now we have to calculate distance travelled in last second that is in 4th second
so,
Snth=u+a/2(2n-1)
S4th=40+(-10)/2(2*4-1)
S4th=40-(5)(7)
S4th=5m
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HOPE IT WILL HELP
PLEASE MARK IT AS BRAINLIEST

Answer 2

u=40m/s

a=10m/s

v=0m/s

now

v=u+at

0=40+(-10)t

t=4s

so total time for journey is 4seconds

now we have to calculate distance travelled in last second that is in 4th second

so,

Snth=u+a/2(2n-1)

S4th=40+(-10)/2(2*4-1)

S4th=40-(5)(7)

=5m

hope it helps plss mark as brainliest