Projectile Motion (2 Viewers)

[ronnknee]

After putting some more thought to it, yea I had to find the relationship between v and u, and sub it into the displacement equation. Instead, I subbed it into the wrong equation. Thanks, I corrected myself

 

[Aerath]

Thanks heaps guys. And I'm just wondering, what's the working out for a question that goes:

A golf ball, at point 0, is hit towards A with an initial speed of V = 12m/s at an angle theta, to the horizontal, just to clear a hedge at A, of height 1m, and distant 4m from 0. If the acceleration due to gravity = 10m/s^2, find the range of values theta, to the nearest degree, at which the ball must be hit so that it will land to the right of A.

Are we meant to find cartesian equation, then sub points in?

 

[lyounamu]

Thanks heaps guys. And I'm just wondering, what's the working out for a question that goes:

A golf ball, at point 0, is hit towards A with an initial speed of V = 12m/s at an angle theta, to the horizontal, just to clear a hedge at A, of height 1m, and distant 4m from 0. If the acceleration due to gravity = 10m/s^2, find the range of values theta, to the nearest degree, at which the ball must be hit so that it will land to the right of A.

Are we meant to find cartesian equation, then sub points in?

Yeah, I did what you thought of doing.

Then I got @ is equal to 81.6188958... degrees or 22.4173475... degrees
And then I sub in those values to see larger angles produce larger x-values because you want larger x values for the ball to land right to the hedge.

 

[Aerath]

I got a cartesian equation of:
y = -5x^2/144*sec^2(@) + x*tan(@)

What number did you sub in?

Edit: Got it.

 

[Aerath]

And there's also this question, which I don't understand at all. If someone could just explain what the diagram is supposed to look like, or something rather, that would just be awesome.

A stone is projected with a velocity of 30m/s from a point at the foot of an inclined plane making an angle of 30* to the horizontal. The path of the projectile can be assumed to be in the vertical plane containing the line of the greatest slope of the inclined plane. If the angle of projection @ > 30*, for what values of @ will the stone strike the inclined plane:
a) horizontally (answer 49*6*)
b) at right angles (answer 70*54*)

g = 10m/s^2

 

[lolokay]

And there's also this question, which I don't understand at all. If someone could just explain what the diagram is supposed to look like, or something rather, that would just be awesome.

A stone is projected with a velocity of 30m/s from a point at the foot of an inclined plane making an angle of 30* to the horizontal. The path of the projectile can be assumed to be in the vertical plane containing the line of the greatest slope of the inclined plane. If the angle of projection @ > 30*, for what values of @ will the stone strike the inclined plane:
a) horizontally (answer 49*6*)
b) at right angles (answer 70*54*)

g = 10m/s^2

I think it would just mean that the stone is thrown up a 30* hill, so you find the angle it would need to be thrown at to hit the hill at an angle of 0* and at a perpendicular angle to the hill; however my answers were a bit off. I got a) as 45* and b) as arctan 3 = 71*34*.

 

[u-borat]

i can confirm both answers are correct.
basically, for part a), you put velocity in the y direction=zero, then make time the subject. Then you sub time into both the x distance and the y distance equations, then you equate these two equations by multiplying the y distance by root 3.
you can do this because in the 30 degree triangle.
with part b), its a bit more complex, and a lot harder to explain...unless there's an easier way. what i did was find the velocites in the x and y direction AS it hits the hill; and again using the 30 degree triangle, you play around with that and an answer pops out eventualy.

basically, the question is in part a; you want the projectile to hit it horizontally, as in....

-/

thats the best diagram i can give, the hyphen is the projectile hitting the hill which is at an angle of 30.

 

[Aerath]

No, that's more than enough help - telling me the working out. Thank you heaps, u-borat, lolokay, lyounamu, Yip, and ronnkee.