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# Projectiles question watch

1. Can someone help here? I am stuck on part (vi).
i)17.2,8ms^-1 ii)1.64s iii)28.2m iv) 0.82s v)3.3m vi)2.72m,No
2. You know that maximum height, h, is always when v=0. So if you know that u = 19 ms^-1, and that a = -9.8 ms^-2, then you can sub into a suvat equation to find h

edit it's part vi. whoops

work out time it takes for the horizontal distance travelled to be 20m, then sub in that time to the vertical component i reckon ?
3. (Original post by Phatty_Magoo)
You know that maximum height, h, is always when v=0. So if you know that u = 19 ms^-1, and that a = -9.8 ms^-2, then you can sub into a suvat equation to find h

edit it's part vi. whoops

work out time it takes for the horizontal distance travelled to be 20m, then sub in that time to the vertical component i reckon ?
Thanks for the reply! I did that as well but what is bothering me is why are we working out the time it takes for the horizontal distance travelled to be 20m? How do you it's not a vertical distance of 20m? Or how do you know if it is not vertical distance+horizontal distance combined which gives a distance of 20m-so like a slanted line with length 20m?

It doesn't say that the direction of the ball's flight is 20m.
4. (Original post by Kalabamboo)
Thanks for the reply! I did that as well but what is bothering me is why are we working out the time it takes for the horizontal distance travelled to be 20m? How do you it's not a vertical distance of 20m? Or how do you know if it is not vertical distance+horizontal distance combined which gives a distance of 20m-so like a slanted line with length 20m?

It doesn't say that the direction of the ball's flight is 20m.
No problem! We want the horizontal distance because it says that the other member is "standing 20m away from Clare", and realistically that'd just be a horizontal distance of 20m as you wouldn't be standing in mid-air (providing that the horizontal ground is flat)

So if you know the time for the horizontal distance to be 20m, then if you sub that time into the right suvat equation to work out how high the ball is at that SAME time when its travelled 20m horizontally

Hopefully that helps !
5. (Original post by Phatty_Magoo)
No problem! We want the horizontal distance because it says that the other member is "standing 20m away from Clare", and realistically that'd just be a horizontal distance of 20m as you wouldn't be standing in mid-air (providing that the horizontal ground is flat)

So if you know the time for the horizontal distance to be 20m, then if you sub that time into the right suvat equation to work out how high the ball is at that SAME time when its travelled 20m horizontally

Hopefully that helps !
Haha Thanks for the help! It makes sense now
6. (Original post by Kalabamboo)
Haha Thanks for the help! It makes sense now

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