Apple thrown from 1.3 m at 2.4 m/s—will it reach 5 m high?

[cherica48]

Help me please I've been stuck on this problem for an hour now...

John throws and apple from a height of 1.3 m with velocity of 2.4 m/s. Will it reach his friend in a treehouse 5 m above the ground?

My book gives the answer "No, it will only reach 1.6 m," but I really want to know how to do this. Any help is greatly appreciated. Thanks!

 

[Jeff Ford]

I would think you'd need to know the angle of the throw and how far away the friend is, but I'm quite a beginner so I'm not sure.

 

[Doc Al]

Consider conservation of energy. Will the apple have enough energy to make it to that height (regardless of the angle it's thrown)?

 

[cherica48]

Whoops, sorry about the homework help violation...in such a rush to figure this out that I didnt see the rule.

I'm pretty sure he wants us to use a Free Fall kinda thing, with an equation like Vf=Vi+2(a)(t), but not all those factors are there. This is basic physics stuff, we're studying like -9.81 acceleration for gravity, and how to find displacement and velocity and all that.

 

[Doc Al]

If you threw the apple straight up at that speed, how high would it go?

 

[cherica48]

If you threw the apple straight up at that speed, how high would it go?

Well that's what the question is asking, and I assume that's displacement. Displacement in our book is

D= .5(Vi+Vf)t

I have Vi(2.4), and Wf would be 0 once it reached the top, but I don't know t. Am I using the wrong formula here? I have no idea what to do...

 

[Doc Al]

You can find the time using the definition of acceleration:
$$v_f = v_i + at$$