Using Projectile motion to find the distance above a wall

[PensNAS]

1. Homework Statement

During a tennis match, a player serves the ball at a speed s, with the center of the ball leaving the racquet at angle θ below the horizontal at a height y0. The net is a distance d away and has a height h. When the ball is directly over the net, what is the distance between them? Express your answer in terms of the given variables and the gravitational acceleration g.



2.Relevant equations

y=dtanθ-d^2(g/(2(scosθ^2))


3.The attempt at a solution

y=dtanθ-d^2(g/(2(scosθ^2))-h

I assumed the equation would give the y at the distance,d. Subtract the height of the net and that would be the distance by which the tennis ball would clear the net. As it turns out, that is not correct. I am a little confused. Any help would be appreciated!

 

[rude man]

Write equations for x(t) and y(t). Tan(θ) plays no part in this. Try sines and cosines instead ... use the x equation to determine the time T it takes for the ball to reach the net and the y equation to determine the height of the ball = y(T) when it reaches the net, then answer = y(T) - h.

 

[PensNAS]

I set x=scos(θ)t and y=ssin(θ)t-0.5gt^2. Solving for t with the x equation I got t=x/scosθ. Plugging that into the y equation I got the same equation of y=dtanθ-d^2(g/(2(scosθ^2)). I tried simplifying that some more and got (2(s^2)sin(2θ)d-gd^2)/2(s^2)cos(θ)^2. It still is not right and I don't know how else to simplify that ungodly fraction:)

 

[rude man]

You got the x equation right and I assume you solved for t correctly (let x = d). Call that t = T.

Your y equation is almost right also, it's missing an initial value of y = y(t=0). When you add it you will have an expression for y(T) where t=T is the time the ball is over the net.

I do owe you one apology: tan(theta) does appear in the y equation after all, since T = d/s*cos(theta). But derive it correctly.

 

[lep11]

By the way, is this high school physics?