A body is thrown with velocity 100m/s, it travels 5m in last second of its upward journey. if the same body is thrown with velocity 200m/s, what distance will it travel in last second of its upward journey?

Question

cp9454 · Answer

A.  5
B.  10
C. 20
D. 25

anonymous · Answer

To me, it 's A

cp9454 · Answer

how, please explain.

anonymous · Answer

nvm, I really don't know how to explain
@tkhunny

tkhunny · Answer

Location: $s_{0}(t) = -4.9t^{2} + V_{0}t + H_{0}$

Location 100: $s_{1}(t) = -4.9t^{2} + 100t + H_{1}$
Location 200: $s_{2}(t) = -4.9t^{2} + 200t + H_{2}$

When does the first one reach its peak?

$t_{Peak1} = -b/(2a) = -100 / -9 = 11.1111111$

$s_{1}(11.111) = -4.9(11.111)^{2} + 100(11.111) + H_{1}$

1 Second sooner

$s_{1}(10.111) = -4.9(10.111)^{2} + 100(10.111) + H_{1}$

This will help you find $H_{1} = H_{2}$

Location 200: $s_{2}(t) = -4.9t^{2} + 200t + H_{2}$

Again, the maximum height is:

$t_{Peak2} = -b/(2a) = -200 / -9 = 22.2222222$

Maximum height is:

$s_{2}(22.222) = -4.9(22.222)^{2} + 200(22.222) + H_{2}$

One second sooner

$s_{2}(21.222) = -4.9(21.222)^{2} + 200(21.222) + H_{2}$

That's about it.  Just walk your way through it, one piece at a time.

tkhunny · Answer

I was bored.  Do you see my errors?

First, -b/(2a) = 100/9.8, not 100/9.

In any case, at some point, the faster one slows to 100 m/s and then follows the same trajectory as the slower one.  Thus, they are the same at the peak.

5 is 5.

anonymous · Answer

Yes!! I am calculating and figure out the mistake there. hihihi Thanks for neat explanation. I got it.