Projectile Motion: Shooting over a hill

[kkernodl]

Homework Statement

A projectile is fired with speed v₀ at an angle θ from the horizontal as shown in the figure (see attachment).

Homework Equations

A. Find the highest point in the trajectory, H. Express the highest point in terms of the magnitude of the acceleration due to gravity g, the initial velocity v₀, and the angle θ.

B. What is the range of the projectile, R? Express the range in terms of v₀, θ, and g.

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

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

E. Find t_g, the flight time of the projectile. Express the flight time in terms of H and g.

The Attempt at a Solution

A. H=(v₀sin(θ))²)/2g

B. R=v₀²sin(2θ)/g

C. I know that arctan(2H/R)=θ, however this isn't the correct answer. I've tried everything I can think of, but don't know how to solve this part.

 

[hotvette]

C. I know that arctan(2H/R)=θ, however this isn't the correct answer

What you need to do is find the time it takes for the range (x) to be R/2 and then find θ such that the height (y) is H.

Your expression is valid for a triangle connecting the starting point to the peak but that doesn't have anything to do with the projectile problem.

 

[kkernodl]

What you need to do is find the time it takes for the range (x) to be R/2 and then find θ such that the height (y) is H.

Your expression is valid for a triangle connecting the starting point to the peak but that doesn't have anything to do with the projectile problem.

I found the position function in the x direction and set it equal to R/2. This gave me that the time, t, such that the position of x is R/2 is t=v₀sin(2θ)/2gcosθ.

I can use trig identities to simplify this to t=v₀sinθ/g

I can find the position function of y: S_y=v₀t-4.9t², but I'm not sure what to plug in.

I'm also very confused as to how I'm supposed to express the answer in terms of H and R, as the question specifically asks.

 

[kkernodl]

I was able to answer it using my original answer.

I had to find the ratio of H/R, which is sinθ/4cosθ or (1/4)tanθ.

I set (1/4)tanθ=H/R and solved for θ, which gave me θ=arctan(4H/R), which is the correct answer.

 

[asteriatic]

D. v0=√(gR/2H)

E. tg=√(2H/g)

I would like to clarify that the equations used in this problem are based on the assumptions of ideal projectile motion, where there is no air resistance and the acceleration due to gravity is constant. In real-life scenarios, there may be factors such as air resistance and varying gravitational forces that can affect the trajectory and flight time of a projectile.

To solve part C, we can use the range equation (B) and substitute in the expression for θ from the arctan equation. This gives us R=v0^2sin(arctan(2H/R))/g. We can then use the trigonometric identity sin(arctan(x))=x/√(1+x^2) to simplify the expression and solve for θ.

Furthermore, it is important to note that the initial speed (v0) in this problem is dependent on the values of H, R, and g. So, if any of these values are changed, the initial speed will also change accordingly.