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 $=5m$