A projectile is fired with speed v0 at an angle from the horizontal as shown in

Question

A projectile is fired with speed v0 at an angle from the horizontal as shown in the figure

I found out that

The highest point in the trajectory (H)= Vo^2 sin(theta)^2 / 2g

Range = Vo cos(theta)* 2Vosin(theta) / g

I need help finding

1 Find the angle ? above the horizontal at which the projectile should be fired. Express your answer in terms of H and R.

2 What is the initial speed? Express v0 in terms of g, R, and H.

3 Find tg, the flight time of the projectile. Express the flight time in terms of H and g.

Explanation / Answer

Express v0 in terms of g,R,H means that when you write your answer for v0, it should be a mathematical expression involving the variables g,R, and H, but no others.

Use trigonometry to break the initial velocity into components:
The initial horizontal velocity is: vx = v cos (theta)
The initial vertical velocity is: vy = v sin (theta)

To calculate the highest height, use conservation of energy on the y-velocity
initial vertical KE = 1/2 m vy^2
= final PE = mgh

Solve for height: h = sqrt (vy^2 / 2g)

vy(t) = initial vy - gt

At the highest point, vy(t) = 0. Solving for that time:
t = vy / g
The total time, therefore is twice that, because what goes up must come down in the same time.
total time = 2 vy / g = 2 v sin (theta) / g

The horizontal position at any time is given by:
x(t) = vx t, so the range is given by
range = v cos (theta) * 2 v sin (theta) / g
= v^2 sin (2 theta) / g

To do the rest of the problem:
Find the vy for which the projectile just barely clears the hill
1/2 m vy^2 = mgh, so vy = sqrt (2gh)
Then find how long it will be aloft
time aloft = 2 vy/g = sqrt (8h/g)

Then find the vx that gives you the desired range.
range = vx t
so vx = range / t = range sqrt (g / 8h)
Find total velocity using pythagorean theorem.
v = sqrt (vx^2 + vy^2) = sqrt (range^2 g / 8h + 2gh)
Find the angle using tan(angle) = vy / vx
angle = arctan (sqrt (2gh) / range sqrt (g / 8h))
= arctan (4 h / range)

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