Projectile Motion Problems: Calculating Maximum Angles and Heights

[hancyu]

i don't get it... so, should i substitute sinO with (1-cosO^2)^1/2 ?

 

[hancyu]

i think theta should be 40.7(degrees)

i got it through trial and error. i just don't know how to solve for it.

 

[Dick]

Good job on getting the approximation. Write the equation as 10*sin(O)-5=2*cos(O). Now put cos(O)=sqrt(1-sin(O)^2). Now square both sides. You get a quadratic in sin(O). Solve for sin(O). Since you squared both sides check that each of your answers actually satisfy the original equation. Now find O.

 

[hancyu]

so 1st i divided the whole equation with -5
i got.
1=2sinO -cosO then, i substituted sqrt(1-sin(O)^2) to cosO
then i got. 1=4sin^2(O) + 1 - sin^2(O) which doesn't make sense.

how should i do it?

 

[Dick]

so 1st i divided the whole equation with -5
i got.
1=2sinO -cosO then, i substituted sqrt(1-sin(O)^2) to cosO
then i got. 1=4sin^2(O) + 1 - sin^2(O) which doesn't make sense.

how should i do it?

You aren't doing the algebra right. If you square 2sinO-cosO you get 4sin^2(O)+cos^2(O)-4*sin(O)*cos(O). There's a cross term you've forgotten and you haven't gotten rid of the square root at all. Get the sqrt part all by itself before you square.

 

[hancyu]

let sinO be x
so its. (2x-1)^2 = 1-x^2

5x^2 - 4x = 0

sin^-1 of .8 = 50+
which is not equal to 40.7(which is a close approximate)

how should i do it

 

[hancyu]

let sinO be x
so its. (2x-1)^2 = 1-x^2

5x^2 - 4x = 0

sin^-1 of .8 = 50+
which is not equal to 40.7(which is a close approximate)

how should i do it

 

[Dick]

You've got the equation wrong now. It's 10*sin(O)-5=2*cos(O). You goofed it up way back when you divided everything by 5. About half your problem is simple sloppiness.

 

[hancyu]

i got it! 104sin^2(O) - 100sin(O) +21 = 0
then O should be 18.05 and 40.7
^^ both satisfies the equations above.

 

[Dick]

I don't think 18.05 works as well. You should check that. But look, if you pay attention and check what you are doing you can solve these problems.

 

[hancyu]

**EDIT**

so... 22/19.15 = t = 1.15s

y=Vyot +1/2gt^2
y=12.0m

vyf^2 - vyo^2 = 2gy

vyf=4.8m/s

is this right?

if the teacher's asking for Vf... how do i get it. since 4.8m/s is Vyf ryt? and Vxf is 19.15. should i use pythagorean theory thinggy?

 

[Dick]

Sure, if you want the total magnitude of the final velocity, use the pythagorean thingy.

 

[hancyu]

help on the first question please. i still can't understand it.

 

[Dick]

Start by rereading post 11. Then tell me what you don't understand.

 

[hancyu]

If you throw a ball upwards at a steep angle it will go away from you and then come back towards you (and hit you on the head if the angle is too steep). If you throw it out at a shallow angle then it always travels away from you and doesn't come back. They want you to find the angle that divides the two cases.

shouldn't there be 2 angles. coz it will be lyk...'between angle1 & angle2 '

 

[Dick]

No. Lyk, coz it will b shallow angle always out, steep angle always starts coming back. There is an angle A such that a<A it will go out, a>A it will come back. Why two angles?

 

[hancyu]

^you said on one of your posts that r(t)^2=x(t)^2+y(t)^2

is that the same with R^2= cos^2 (t)+ sin^2 (t)

what should i multiply cos and sin with?

 

[Dick]

No. Do this the same way you did the other problems. x(t)=v*cos(A), y(t)=v*sin(A)-(1/2)gt^2. v is the initial velocity, A is the angle from the horizontal. Yes, r(t)^2=x(t)^2+y(t)^2. You have to figure out a condition on r(t)^2 that will insure it's always increasing. I.e. (r(t)^2)' is always positive. I.e. (r(t)^2)' is never zero. I've mentioned before that this is not an easy problem. Are you sure you need the extra credit that much?

 

[hancyu]

yes. i really need extra credit that bad.

Vo is not stated in the problem...what should i do?

R is range ryt? then x(t) is the x component and y(t) is the y component?

 

[Dick]

Leave Vo unknown. It doesn't matter. It will cancel out. Yes, x(t) and y(t) are the components. I don't know what R is. r(t) is the distance from you to the projectile at time t.

 

[hancyu]

r(t)^2 = vcos(t)^2 + [vsin(t) - 1/2(gt^2)]^2

then what?

 

[Dick]

You really can't just keep asking me questions without putting any thought in before hand, ok? I already told you. You have to make sure that the derivative of r(t)^2 is never zero. Then it can't come back. And I wouldn't use t for both time and angle, it's going to be confusing. Differentiate it, ok? With respect to t. Not t. Told you it would be confusing.

 

[lorenzom21]

No, it's not 45 degrees. If the launch angle is theta then you work out x(t) and y(t). The distance squared from the launch point (0,0) is r(t)^2=x(t)^2+y(t)^2. You want to take the derivative of r(t)^2 with respect to t and then figure out for which angles theta that expression has a zero derivative. These are the ones where it comes back towards you.

Dick, why is it that I have to find the derivative of r(t)^2? the distance is r(t) so shouldn't I derive r(t) with respect to t and not r(t)^2?

 

[Dick]

Dick, why is it that I have to find the derivative of r(t)^2? the distance is r(t) so shouldn't I derive r(t) with respect to t and not r(t)^2?

If r(t) is increasing/decreasing then r(t)^2 is increasing/decreasing. You can do either one. I just suggest r(t)^2 because it's easier. It doesn't have a square root in it.