Projectile Motion question help

[DisaFear]

Hello

First questions are always nooby questions, dont kill me yet

What is the angle of projection? What does the "v" stand for? Final velocity? That doesn't help much here, does it. It's a symmetric projection anyway, we can work that put from the initial speed?

[xZero]

since your given the maximum height 4m, lets focus on the vertical velocity which is 10sin(x) where x is the angle of the projection. Use v^2 = u^2 + 2ax, where x = 4m, v = 0ms^-1 (the vertical velocity at the max height is 0), u = 10sin(x) and  a = -9.8ms^-2. Plug all the numbers in and solve for x

[DisaFear]

Plugged it into the CAS, gives a horrible answer

As seen on wolfram alpha

Cant be that complex...obv i am doing something wrong?

[xZero]

I got 62.3 degrees

[DisaFear]

So smart, correct
Since plugging numbers in a calc doesnt cut it, may you please give me an explanation?

[xZero]

I plugged the numbers into my cas, solve for x and got 4 solutions, just look for the positive acute angle. Or if your using a scientific calc, 78.4 = 100sin^2(x). 0.784 = sin^2(x). 0.8854 = sin(x). x = 62.3 degrees

[cohen]

Since it's symetrical, couldn't we use R = (V^2 sin(2*theta))/g , and then just work out sin(theta) from this?

[xZero]

what equation is that :S (I only know the 5 constant acceleration formulas)

[onur369]

ive been staring at this question for 20mins, I cant do it, it doesnt look possible with the equations i have :/

[DisaFear]

Ok, let's see...

When I do sin(62.3) the cas gives -0.51
When I do sin(62.3) with a degree symbol, gives 0.8854 (makes sense right)
When I do solve(sin(x)=0.8854,x), it gives 6.28319*n1 + 2.05 or 6.28319*n1 + 1.08
Using the degree signs with x does not work. Do I need to change modes or something?

I have gotten to where have gotten in your calcs, but this final step of calculations i cannot get past. What is n?

[xZero]

Hint: Change your cas to degrees mode, right now the answer you're getting is in radians

Also, have you done the general solution for trig? If so that's what the n is, if not well you'll understand when you get to it in method

Edit: @schnappy ahhh rofl I just use x = ut for horizontal distance

[schnappy]

what equation is that :S (I only know the 5 constant acceleration formulas)

If you plug in v*cos(theta) in to an equation for the time (s=d/t) which comes from the vertical, it boils down to x(hori)=v*v*sin(2*theta)/g

Lets just look at the vertical:
vv=uu+2ax
x=4
u=10sin(theta)
v=0
a=-10
0=100sin^2(theta) -80
sqrt(0.=sin(theta)
inverse sine of sqrt(0., from 0-90deg and bam answer.

Use a scientific calculator if you can't use a CAS properly.

[DisaFear]

Ok, scientific calc worked, i changed the mode to degrees on cas doesnt work
mhmhmhm shrug

[cohen]

@xZero, i don't know exactly where the equation comes from. But my teacher gave us some help book (i'm assuming its from the old study design) it states:

Code: [Select]

for flights without air resistance, that start and end at the same height, the following formula for the range can be used:
R = (v^2 sin(2 theta))/g
where R is th range, v is the initial speed and theta is the angle of projection)

I just reread the question, and remembered that the intial speed is u=10sin(x), so unfortunately we can't use the formula :/ (Unless you're willing to do all the work involving sin^2(x)

[schnappy]

If you take the variables out and use pronumerals... this is where the range equation comes from:
http://vce.atarnotes.com/forum/index.php/topic,39785.msg417898.html#msg417898