Projectile Motion, LAST QUESTION! (1 Viewer)

[lyounamu]

This is the last question from the Fitzpatrick 3 Unit book on Projectile Motion.

This question sounds really simple but I don't know what to substitute and what formula to use. I already used the cartesian equation and the maximum height formula but they don't seem to work.

Here we go:

Find the speed and direction of a particle which, when projected from a point 15 m above the horizontal ground, just clears the top of a wall 26.25 high and 30 m away.

Thanks for the help in advance.

EDIT: I need help with the second part of this question too. Sorry!

A particle projected from a point meets the horizontal plane through the point of projection after travelling a horizontal distance a, and in the course of its trajectory attains a greatest height b above the point of projection. Find the horizontal and vertical components of the velocity in terms of a and b. Show that when it has descirbed a horizontal distance x, it has attained a height of 4bx(a-x)/a^2.

I alredy proved that the horizontal component of velocity is a/2 . square root of (g/2b) and the vertical component being square root of (2gb) where g = gravity via manipulation of the maximum height formula and the range formula. I just need help with the 2nd part to prove the height.

 

[minijumbuk]

http://community.boredofstudies.org/showthread.php?t=176678

 

[lyounamu]

http://community.boredofstudies.org/showthread.php?t=176678

Awesome!

Thanks for the help. I appreciate it!!

EDIT: ANYONE at Q2?

 

[lolokay]

A particle projected from a point meets the horizontal plane through the point of projection after travelling a horizontal distance a, and in the course of its trajectory attains a greatest height b above the point of projection. Find the horizontal and vertical components of the velocity in terms of a and b. Show that when it has descirbed a horizontal distance x, it has attained a height of 4bx(a-x)/a^2.

I alredy proved that the horizontal component of velocity is a/2 . square root of (g/2b) and the vertical component being square root of (2gb) where g = gravity via manipulation of the maximum height formula and the range formula. I just need help with the 2nd part to prove the height.

just substitute the usual expressions for max height, max distance, and distance into the expression and verify that it is equivalent to the expression for height. or do that and then work backwards. I'm not sure which side you work with in show questions

 

[lyounamu]

just substitute the usual expressions for max height, max distance, and distance into the expression and verify that it is equivalent to the expression for height. or do that and then work backwards. I'm not sure which side you work with in show questions

Yeah. I did that and I got it right. But the working out was sooooooo long. It was extremely long. I guess that's the only way, (if not the best) to solve this question.

I just wanted to see how people did this question because my working out was not an efficient one.

Thanks a lot for that confirmation though.

 

[lolokay]

it didn't seem like that long working to me

edit: did you use..

[s = sin@, c = cos@]

b = v^2 s^2 / 2g

a = 2 v^2 sc / g

x = v c t

y = v s t - 1/2 * g t^2

then just substitute into 4bx(a-x)/a^2. ?

 

[lyounamu]

it didn't seem like that long working to me

edit: did you use..

[s = sin@, c = cos@]

b = v^2 s^2 / 2g

a = 2 v^2 sc / g

x = v c t

y = v s t - 1/2 * g t^2

then just substitute into 4bx(a-x)/a^2. ?

Ok, I will try that. Thanks.

EDIT: Just realised that I wrote my working out in an extremely elongated way. I could have written one line instead of unnecesary 10 lines. Thanks for the help.