# Using Projectile motion to find the distance above a wall

• PensNAS
In summary, 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. To find the distance between the ball and the net when the ball is directly over the net, one must use the equations x(t) = s*cos(θ)*t and y(t) = s*sin(θ)*t - 0.5*g*t^2, with the initial value y(0) = y0. Solving for t using the x equation and plugging it into the y equation, we get an expression for y at time t when the
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!

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.

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:)

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.

By the way, is this high school physics?

## 1. How do you calculate the distance above a wall using projectile motion?

To calculate the distance above a wall, you will need to use the equation D = V0^2 * sin(2θ)/g, where D is the distance, V0 is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity. This equation assumes a projectile launched from ground level.

## 2. What information do I need to know in order to use projectile motion to find the distance above a wall?

In order to use projectile motion, you will need to know the initial velocity of the projectile, the angle at which it is launched, and the acceleration due to gravity. You may also need to consider factors such as air resistance and the height of the wall.

## 3. Can I use projectile motion to find the distance above a wall if the wall is not perfectly vertical?

Yes, you can still use projectile motion to find the distance above a wall if it is not perfectly vertical. However, you will need to take the angle of the wall into account when calculating the angle of projection and the initial velocity.

## 4. Is there a limit to the distance that can be calculated using projectile motion?

No, there is no limit to the distance that can be calculated using projectile motion. However, the accuracy of the calculation may decrease for extremely large distances, as other factors such as air resistance may come into play.

## 5. How can I use projectile motion to find the distance above a wall in real-life situations?

In real-life situations, you can use projectile motion to find the distance above a wall by measuring the initial velocity and angle of projection of the projectile, and then plugging those values into the equation D = V0^2 * sin(2θ)/g. You may also need to make adjustments for factors such as air resistance and the height of the wall.

Replies
39
Views
3K
Replies
11
Views
1K
Replies
12
Views
404
Replies
42
Views
1K
Replies
2
Views
3K
Replies
19
Views
2K
Replies
1
Views
2K
Replies
3
Views
1K
Replies
8
Views
2K
Replies
3
Views
4K