How Do You Calculate Projection Speed and Maximum Height in Projectile Motion?

[kfox1984]

i have to answer this question for an assignment that I need to do for mechanics. I am really really stuck - would somebody please mind helping...

A ball thrown at an angle α to the horizontal just clears a wall. The horizontal and vertical distances to the top of the wall are X and Z respectively, with X.tan α>Z. Show that the projection speed is...

X/cos α { g/(2(X.tanα – Z)}^1/2

Where g is acc due to gravity. Determine the position of the highest point reached in terms of X. Z and tan α.

 

[Doc Al]

Hint: Write equations for the vertical and horizontal positions as functions of time.

 

[kfox1984]

Thanks - I have already done that but still nothing...

 

[Doc Al]

Thanks - I have already done that but still nothing...

Show what you did. Make use of the fact that when x = X, y = Z.

 

[kfox1984]

I have done that -
i'm not sure which equation to use...

I know that tx = ty so i have tried rearranging things and substituting. but i still have v's left in everything...

i don't understand where the tan has come from...

I'm guessing that i have to use v^2 = u^2 + 2as as the final eqn has a 2 in it but i don't understand how the two has got to the bottom of the fraction sign. I also don't understand where the squre root has come from...

 

[kfox1984]

I don't even know if I'm supposed to be using calculus or not - this is supposed to be an as level mechanics problem

 

[Doc Al]

Again I ask you to write down the expressions for horizontal (x) and vertical (y) positions as a function of time. (They will also involve V and α.) The tan and the 2 will fall into place automatically when you later set up and solve those equations.

(Nothing to do with calculus, just a little algebra.)

 

[kfox1984]

xcos(α) = (U + V)/2 * t , zsin(α) = (u + V)/2 * t

 

[kfox1984]

or x = [(ucos(α) + V/)2] *t and z = [(usin(α) + V)/2] * t

 

[Doc Al]

xcos(α) = (U + V)/2 * t , zsin(α) = (u + V)/2 * t

Those aren't correct. Review the kinematic equations here: https://www.physicsforums.com/showpost.php?p=905663&postcount=2"

(The ones you want relate displacement and time.)

or x = [(ucos(α) + V/)2] *t and z = [(usin(α) + V)/2] * t

Getting closer, but still not right.

 

[kfox1984]

I'm sorry but i still really don't understand what i am supposed to do with these.

 

[Doc Al]

Start with this general equation:
$$x = x_0 + v_0 t + (1/2) a t^2$$

How would you apply it to the horizontal and vertical motion of your projectile?

 

[kfox1984]

ok so s = ut + 1/2at^2...

x = ucos(α)t , Z = usin(α)t 1/2gt^2 but I'm still unsure about what i am supposed to do next...I already tried rearranging the X one in terms of t then substituting it into the other...which didn't work...if i divide them then i loose the cos(α) using pythagoras was still wrong...?

 

[kfox1984]

sorry that's supposed to be z = u.sin(α) - 1/2gt^2

 

[kfox1984]

i've made a substitution...

x/ucos9α) = t

Z = u.sin(α).(x/cos(α)) - 1/2g[(x/ucos(α))^2]

is this correct?

 

[Doc Al]

I already tried rearranging the X one in terms of t then substituting it into the other...which didn't work...

Try it again. (It worked for me. ) Take the X equation and solve for t, then substitute that into the Z equation.

sorry that's supposed to be z = u.sin(α) - 1/2gt^2

Good. But you left out a t in the first term on the right hand side.

 

[Doc Al]

i've made a substitution...

x/ucos9α) = t

Z = u.sin(α).(x/cos(α)) - 1/2g[(x/ucos(α))^2]

is this correct?

Almost. You dropped a u in the second term. But you are on the right track.

 

[kfox1984]

i get -

Z = usinα.(x/cosα) - 1/2g[x/ucosα]^2 but now i have a cos^2 which I'm not sure how to get rid of??
Thanks

 

[kfox1984]

re arranging i get -

u^2 = xthanα - 1/2.g.(x^2/zcos^2α)

Which is still incorrect, i really don't know where i am going wrong sorry

 

[Doc Al]

i get -

Z = usinα.(x/cosα) - 1/2g[x/ucosα]^2 but now i have a cos^2 which I'm not sure how to get rid of??
Thanks

(1) Fix that second term--you left out a "u".
(2) You don't want to "get rid of the cos^2". Just expand that square.
(3) Simplify as much as possible, then solve for u. (Hint: Solve for 1/u^2 first.)

 

[kfox1984]

i've done it...Thankyou soo much for your help. can I please just ask one last question;

finding the highest position i use the fact that v = 0 in the y dirn, is that correct, and i do the same thing pretty much substituting t again?

 

[Doc Al]

i've done it...Thankyou soo much for your help.

Good! And you're welcome.

can I please just ask one last question;

finding the highest position i use the fact that v = 0 in the y dirn, is that correct, and i do the same thing pretty much substituting t again?

Yes, use the fact that the vertical component of speed is zero. Find the time when that takes place. Then just plug that time into your equation for y.