Can someone please help me with Physics??

Question

The questions asks "For what launch angle will the height and range of a projectile be equal?" I have spent over 2 hours trying to figure this question out, and the furhest I have been able to get is cos (x) = (-.5)/(v-initial)

Answer 1

Break the initial velocity into components

vy = v sin theta
vx = v cos theta

The time aloft is given by:
t = 2vy / g = 2v sin theta / g

Range = vx t = 2v^2 sin theta cos theta / g

To get the height, use conservation of energy on just the vertical velocity

1/2 mvy^2 = mgh
h = (v sin (theta))^2 / 2g

h = r
(v sin (theta))^2 / 2g = 2 v^2 sin theta cos theta / g
tan theta = 4
theta = arctan (4) Grab your calculator.

Answer 2

Eh just draw it out. 10 cm for the range, 10 cm for the height.

Draw, the curve, Get a protractor and measure the angle, like ?

Answer 3

The formula for range is:

r = v_i²sin(2θ)/g

The formula for height is:

h = (v_i•sin(θ))²/(2g)

Set them equal:

v_i²sin(2θ)/g = (v_i•sin(θ))²/(2g)

Divide by v_i² and multiply by g, and those two variables magically disappear:

sin(2θ) = sin²(θ)/2

Now use the trig identity: sin(2θ) = 2sinθcosθ

2sinθcosθ = sin²(θ)/2

Divide both sides by sinθ:

2cosθ = (sinθ)/2

Multiply by 2 and divide by cosθ:

4 = sinθ/cosθ = tanθ

That means θ = arctan(4).