Projectiles Questions (1 Viewer)

[iSam]

Just a few quick questions:

A child hurls a stone horizontally off a breakwall 9.8m above the water at 24.5ms^-1.
a) How long will it take to hit the water?
b) How far will it travel horizontally?
c) What will be its horizontal impact velocity?
d) What will be its vertical impact velocity?
e) What will be its net impact velocity and at what angle?

How would I go about answering these?

 

[香港!]

Just a few quick questions:

A child hurls a stone horizontally off a breakwall 9.8m above the water at 24.5ms^-1.
a) How long will it take to hit the water?
b) How far will it travel horizontally?
c) What will be its horizontal impact velocity?
d) What will be its vertical impact velocity?
e) What will be its net impact velocity and at what angle?

How would I go about answering these?

a)s=ut+(1\2)at²
-9.8=24.5(t)-4.9(t)² taking a=-9.8
4.9t²-24.5t-9.8=0
t=[24.5+-sqrt(792.33)]\9.8
=5.37s ignoring the negative value.
b)Consider horizontal component...
s=ut=24.5(5.37)=131.565m
c)horizontal component remains the same throughout the projection.
so horizontal impact velocity is v=24.5m\s
d)
vertical impact velocity v=u+at=(0)-9.8(5.37)=-52.626 m\s
e)net impact velocity would be obtained by using pythagorus:
V=sqrt [(-56.626)²+(24.5)²]=58.049 m\s
to get angle, tan @=(52.626)\(24.5)
@=65 degrees to da horizontal...

I haven't checked these and i'm inefficient using these physics formulas to do them... so could be wrong... hehehe

 

[exa_boi87]

I confirm the above answers ..

 

[iSam]

The answer to a) should be 1.414 or root 2. I got that answer using the maths method (quadratic formula and everything), but when I use the physics equations I get the same answer as you guys.

 

[tennille]

For a), you use s= ut + 1/2at^2.
ut = 0, because the initial velocity is 0.
s=9.8, a=9.8,

So you get, 9.8 = 0 + 1/2*9.8 * t^2
t^2 = 2

Therefore t=1.414 sec.

The whole initial and final velocity is confusing, but that's how you get the right answer for this, even though technically the final velocity is 0.

 

[exa_boi87]

Dammit ! .. thanks for clearing that up, of course as the initial velocity is horizontal, the vertical component must be 0, and as we are determining time taken for the object to reach a vertical destination, we would take into consideration this value for u[y].

Cheers

 

[iSam]

Thanks

Thanks for all your help guys.