Projectile Motion- What angle does range equal max height?

[Acnhduy]

Homework Statement

At what projection angle will the range of a projectile equal its maximum height?

Homework Equations

Big 5

v_f²= v_i² + 2ad

The Attempt at a Solution

So I attempted to solve this question by randomly subbing in equations into each other.

If i draw out a diagram to solve for the angle, I would need the x and y components.

Let V equal the total velocity.

v_iy= Vsinθ
Plugging into v_fy²= v_iy² + 2a_yd_y , I get

v_f²= V²sin²θ + 2a_yd_y

Now I isolate for d, and end up with d_y= (V²sin²θ)/2ay

Ok, after I got the displacement vertically, since I am trying to find the point where they both are at equal magnitude, I now solve for displacement horizontally.

Now using the range formula, I can determine the horizontal displacement.

d_x= (V²sin2θ)/g

(g=a_y)
With these two formulas, I substitute and...

(V²sin²θ)/2a_y = (V²sin2θ)/a_y

sin²θ = a_ysin2θ

sin²θ = a_y(2sinθcosθ)

sinθ = 2a_ycosθ

sinθ/cosθ = 2a_y

tanθ = 2(-9.8)

θ = tan^-1(-19.6)

θ = 87°

However... this is not the answer that the answer key states. The answer shown is 76°, and I do not know where I have gone wrong, or if I am doing this right.

Sorry if it was a mess and confusing, thanks for the help!

 

[voko]

Homework Statement

At what projection angle will the range of a projectile equal its maximum height?

Homework Equations

Big 5

v_f²= v_i² + 2ad

The Attempt at a Solution

So I attempted to solve this question by randomly subbing in equations into each other.

If i draw out a diagram to solve for the angle, I would need the x and y components.

Let V equal the total velocity.

v_iy= Vsinθ
Plugging into v_fy²= v_iy² + 2a_yd_y , I get

v_f²= V²sin²θ + 2a_yd_y

Now I isolate for d, and end up with d_y= (V²sin²θ)/2ay

No. That should be $$ d_y = - \frac {v^2 \sin^2 \theta} {2a_y} $$ Note the minus sign.

Ok, after I got the displacement vertically, since I am trying to find the point where they both are at equal magnitude, I now solve for displacement horizontally.

Now using the range formula, I can determine the horizontal displacement.

d_x= (V²sin2θ)/g

(g=a_y)
With these two formulas, I substitute and...

(V²sin²θ)/2a_y = (V²sin2θ)/a_y

Assuming the above is correct (which it is, because you made another sign error there!), how do you get this:

sin²θ = a_ysin2θ

?

 

[Acnhduy]

I got sin²θ = a_ysin2θ

from

(V²sin²θ)/2a_y = (V²sin2θ)/a_y

Cross multiply and cancel out one of the a_y

a_y(V²sin²θ) = 2a_y(V²sin2θ)

[STRIKE]a_y[/STRIKE] (V²sin²θ) = [STRIKE]2 [/STRIKE]a_y(V²sin2θ)

Divide both sides by V²

[STRIKE]V²[/STRIKE]sin²θ = a_y([STRIKE]V²[/STRIKE]sin2θ)

sin²θ = a_ysin2θ

 

[voko]

You need to brush up on your algebraic manipulation skill. You do not cancel "one of the a" like that.

 

[Acnhduy]

If one side has "2a" and the other has "a", can't you divide out one of the "a" by dividing both sides by it?

 

[Acnhduy]

Ohhh

 

[voko]

If you divide both sides by "a", each side gets divided by "a". So if one side has "2a", and the has "a", how can the result of that be "a" and "1"? Do you really think that the inverse operation, "a" times "a", is "2a"?

 

[Acnhduy]

Ok I see what you mean. I'm left with 2 instead of 'a'. Mistakened as a square

 

[voko]

So what do you get now?

 

[Acnhduy]

Thanks for the help :). After correcting that I end up with 76° :D