1,000,000 Question Thread :D

[Aden]

Sorry, I stand corrected. If you know the horizontal and vertical components, then you can find the actual velocity of the object at that time.

As for the bouncing ball, it isn't the weight force that is changing (unless it suddenly experiences a different gravitational force), but the normal force that can change when it hits the ground and slows down.

EDIT: For your question, remember that the vertical component is under the influence of gravity (). So after one second, the vertical velocity component would be: , and after two seconds: .

[kenhung123]

Oh ok thanks, but how come it doesnt work with my method? Is v not the instantaneous velocity?

[Aden]

Your method doesn't work since you used 28m/s as the intial velocity in your v=u+at formula. The cricket ball is shot out of the cannon thingy at 28m/s, and is the resultant velocity from the horizontal and vertical components (you even showed it youself by using trigonometry and imagining the triangle). However, what you have to realize is that 28m/s is NOT the vertical component and so does NOT undergo an acceleration of -9.8m/s. The 28m/s info is only useful at the very beginning, when you're trying to find the initial velocities of the horizontal and vertical components.

You already found the vertical component in question 4a (14m/s), use that value as your initial velocity in v=u+at.

[kenhung123]

Thanks

Having trouble with e

From d I got 11.4m/s so I just used the formula v=x/t since v=11.4 and t=1 x=11.4m?

[Aden]

You will need to find the vertical displacement as well (you have the horizontal displacement as 11.4m). Once you've found that, use pythagoras to find the resultant displacement.

[kenhung123]

11.4m/s is the resultant velocity for 9d though?
The vertical component is 1.5m/s and horizontal 11.3m/s
Do you mean (1.5^2+11.3^2)^0.5?

[Aden]

Yes, I meant 11.3 (whatever you got for the horizontal displacement). I don't think your vertical component is correct though, use , you already know the initial vertical veolocity, time and acceleration (-9.8m/s).

[kenhung123]

Oh Aden, in heinemann it said the only way to obtain horizontal component is using x=vt. Why is that?

[Aden]

I don't think they meant that it is the 'only' way of getting the horizontal displacement, it is just the 'easiest' way of doing it. This is because the horizontal velocity remains constant, so you don't need to use any other method.

[kenhung123]

For 6a I did x=ut+1/2at^2
u=5, t=?, a=9.8, x=2

Gave t as 0.31s

For the other ball I did
u=7.5, t=?, a=9.8, x=2

Gave t as 0.23s

[Lighties]

If you think about it in a picture, the ball is 2m in the air, and it's hit horizontally. u=5 is for the horizontal component, not the vertical, and thus the distance travelled wouldn't be 2m. Similarly with the other ball.

Therefore you use the vertical component. Then u=0, x=-2, a=-9.8. I think I worked the time out to be for both balls.

Not sure if I'm right though. =S

[kenhung123]

Right, that is a good point so if down is negative accelerations then x down would be negative displacement?

[Lighties]

Yep. (:

Just state one direction (eg. up) as positive, and the other direction will always be negative, for that question.

[kenhung123]

Thanks

Can you tell me what I did wrong in the question D? (I got A and B correct using trig, which were 4 and 6.9m/s)
c) u=6.9m/s, v=0, a=-9.8, t=?
Using v=u+at
        0=6.9-9.8t
        t=0.7s
d)v=0, u=6.9, a=9.8, x=?
Using v^2=u^2+2ax
         0=6.9^2+2*-9.8*x
          x=2.43m

[Aden]

Which question is this for?