Question

a body is thrown vertically upward with 45 m/s .distance travelled by the body in 5th second is

Answer 1

 Given,
Let g = 10m/s^2 
      u = 45m/s 
5th second is from t=4s to t=5s
 At,t=4s
The distance at t=4s
S=ut-1/2*g*t²  = (45*4) - 1/2 * 10 * 4*4  = 180 - 80  = 100m
And velocity at t=4sv=u-gt  = 45 - (10*4)  = 5m/s
Similarly,t=5s
The distance at t=5sS=ut-1/2g(t²)
=(45*5)-1/2*10*5*5 
 = 225-125  = 100m
But this seems that distance travelled is (100-100)= 0m 
Velocity at t=5s
 v=u-gt  = 45 - 10*5 = -5m/s
It is clearly seen that velocity at t=4s has just been retarded at t=5s,which means that at t=4.5s the velocity must have been = 0 m/s
Hence distance should be travelled  till t=4.5s, which is
S=ut-1/2g(t²)  = (45*4.5)- 1/2*10*4.5*4.5  = 202.5 -  1/2*10*20.25  = 202.5 - 101.25  = 101.25m
The answer is 101.25m 

Answer 2

Answer:

1.25

Explanation:

distance travelled by the body 5th sec