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# Mechanics - motion in a straight line watch

1. An object is thrown vertically downwards with speed V. During the sixth second of its motion it travels a distance h. Find V in terms of h and g.

This is my working:

t=6 v=V u=0 g=a s=h

s=vt -1/2 at^2

h=6v -1/2 xgx6^2
h= 6v - 18g

6v = h+18g so v = (h+18g)/6

This is NOT the right answer. And help would be received gratefully. :-)
2. (Original post by Zarantulas)
An object is thrown vertically downwards with speed V. During the sixth second of its motion it travels a distance h. Find V in terms of h and g.

This is my working:

t=6 v=V u=0 g=a s=h

s=vt -1/2 at^2

h=6v -1/2 xgx6^2
h= 6v - 18g

6v = h+18g so v = (h+18g)/6

This is NOT the right answer. And help would be received gratefully. :-)
If it is thrown downwards with speed V then the initial speed (u) will be V.

It looks like you've mistaken V for the final speed instead
3. By that calculation, V =(h-18g) /6

Unfortunately this is still not the textbook answer but thanks anyway!
4. (Original post by Zarantulas)
By that calculation, V =(h-18g) /6

Unfortunately this is still not the textbook answer but thanks anyway!
The object travels a distance h during the 6th second, it hasnt travelled a distance h after the sixth second of motion.
Does that help?

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