projectile motion

Hi have you managed to answer the question? If not I'll try and offer some help.

You have to split the motion of the ball to vertical and horizontal components remembering that acceleration is independent of the horizontal component (and so horizontally a=0). The question also states that the ball was projected horizontally and so vertical initial velocity is zero. Splitting the motion horizontally and vertically, the acceleration needs to calculated using the vertical component only. For part c, the question asks for the speed. Speed is a scalar quantity and so you need to find the velocity of the horizontal and vertical components then find the resultant using Pythagoras. Hope this helps.

Was there a diagram for this question?

Is the ball being projected along the slope of the board, or horizontally so that is undergoes free-fall until it hits the board again? From (b), I imagine it's the first one, but I'm not sure.

Hi have you managed to answer the question? If not I'll try and offer some help.

You have to split the motion of the ball to vertical and horizontal components remembering that acceleration is independent of the horizontal component (and so horizontally a=0). The question also states that the ball was projected horizontally and so vertical initial velocity is zero. Splitting the motion horizontally and vertically, the acceleration needs to calculated using the vertical component only. For part c, the question asks for the speed. Speed is a scalar quantity and so you need to find the velocity of the horizontal and vertical components then find the resultant using Pythagoras. Hope this helps.

Yeah I've attached it.

Oh. Haha, a combination of both then :s

You've done part (a), so for part (b):

You must remember acceleration is a vector quantity, so you must do it separately across the board and down the slope of the board. Across the board, the acceleration is zero, but down the board it is not. So:

s=1.2m
u=0 m/s
v
a=?
t=0.9s

So,
$[br]s = ut + \dfrac{1}{2}at^2[br]$

$[br]a = \dfrac{2(s - ut)}{t^2}[br]$

Putting the numbers in gives you 2.96m/s².

Oh. Haha, a combination of both then :s

You've done part (a), so for part (b):

You must remember acceleration is a vector quantity, so you must do it separately across the board and down the slope of the board. Across the board, the acceleration is zero, but down the board it is not. So:

s=1.2m
u=0 m/s
v
a=?
t=0.9s

So,
$[br]s = ut + \dfrac{1}{2}at^2[br]$

$[br]a = \dfrac{2(s - ut)}{t^2}[br]$

Putting the numbers in gives you 2.96m/s².

How did you get s=1.2?

The length of the board in metres (because I'm resolving down the length of the board, since acceleration is a vector).

Can you help me with part c please?

So, they want the speed (not the velocity), which means you can ignore all the vectors stuff. Write down what you have (you've got the distance, s, initial speed, u, time, t, and you want the final speed, v). Then see which SUVAT equation you need to use .

So, they want the speed (not the velocity), which means you can ignore all the vectors stuff. Write down what you have (you've got the distance, s, initial speed, u, time, t, and you want the final speed, v). Then see which SUVAT equation you need to use .

Hi have you managed to answer the question? If not I'll try and offer some help.

You have to split the motion of the ball to vertical and horizontal components remembering that acceleration is independent of the horizontal component (and so horizontally a=0). The question also states that the ball was projected horizontally and so vertical initial velocity is zero. Splitting the motion horizontally and vertically, the acceleration needs to calculated using the vertical component only. For part c, the question asks for the speed. Speed is a scalar quantity and so you need to find the velocity of the horizontal and vertical components then find the resultant using Pythagoras. Hope this helps.

Thank you.

Sorry to keep bothering you, but when your working out anything to do with the vertical is U always 0?

No problem. To answer your question above, the vertical initial velocity is not always zero but it is in this case because the ball is projected horizontally.

Remember to find speed you have to find the final velocity for both the horizontal and vertical components and find the resultant using Pythagros.