A body is thrown upwards with velocity 100 M per second and it travels 5 M in the last second of its upward journey if the same body is thrown upward with velocity 200 m per second what distance will it travel in the last second of its upward journey?

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Answer 1 · 2017-12-31T16:33:21

The distance travelled in the last second is #=5m#

The greatest height is attained when the velocity #v=0#

Applying the equation of motion

#v=u+at#

#u=100ms^-1#

#a=-g=-10ms^-2#

Therefore,

The time to reach the greatest height

#t=100/10=10s#

Applying the equation of motion to get the height as a function of time

#h(t)=ut+1/2at^2#

#h(t)=100t-1/2g t ^(2) =100t-5t^2#

Therefore,

The distance travelled in the last second is

#h(10)-h(9)=100*10-5*10^2-100*9+5*9^2=5m#

When #u_1=200ms^-1#

#t_1=200/10=20s#

#h_1(t)=200t-5t^2#

#h(20)-h(19)=200*20-5*20^2-200*19+5*19^2#

#=5m#

The distance travelled in the last second is #=5m#