.Anobjectdroppedfromaclifffallswithaconstantaccelerationof10m/s2.Findits speed 5 s after it was dropped. {remember dropped means 0 initial velocity}.
How high was the cliff, if the object hit the ground after 8 second?
Answers
Answered by
0
First body initial speed u=0
Final speed =3m/s
Second body initial speed=4m/s
Body is dropped from height h
So a=g=10m/s
2
v
2
=u
2
+2gh
3
2
=0+20h
9=20h
h=
20
9
Second object is also dropped from same height.
therefore, h=
9
20
u=4m/s
v
2
=u
2
+2gh
v=5m/s
please follow me and mark my answer in brain list ♥️ ♥️♥️♥️
Answered by
2
Answer:
s = ut + 1/2at^2
= 0m/s * 5sec + 1/2 * 10m/s^2 * (5)^2seconds
= 1/2 * 10m/s^2 * 25sec
= 125m
distance = 125m
speed = distance/time
= 125m / 5s
= 25m/s
s = ut + 1/2 at^2
= 0 * 8 + 1/2 * 10m/s^2 * 8^2
= 1/2 * 10m/s^2 * 64
= 320 m
height = 320 m
hope it helps you
Similar questions