How Should a Quarterback Throw to Hit a Stationary Receiver?

[Natko]

Homework Statement

A quarterback is running across the field, parallel to the line of scrimmage, at a constant speed of 2.5m/s, when he spots an open, stationary receiver straight downfield from him (ie, in a line parallel to the sidelines). If he can throw the football at a speed of 8.0m/s, relative to himself, at what angle, relative to the sidelines, must he throw it in order to hit the receiver? How far downfield was the receiver?

Homework Equations

Pythagorean theorem
SOH CAH TOA

The Attempt at a Solution

 

[haruspex]

Your diagram performs the vector sum of two velocities at right angles. You sum two velocity vectors if one of them represents a velocity relative to the other. But the relative velocity is not given here - it's the thing you are trying to find.
See if you can get the right diagram.

 

[Natko]

Does this look better?

Sorry, it's flipped on PF even though I rotated it on my computer.

 

[haruspex]

Does this look better?

Yes, except that the arrow on the 8m/s goes the wrong way. When adding vectors diagrammatically, the contributing vectors connect nose-to-tail; the resultant runs from the tail of the first to the nose of the last.

 

[HalfLostAndSearching]

To solve this problem, we can use the Pythagorean theorem and the concept of vector motion. Since the quarterback is running parallel to the line of scrimmage, his velocity can be represented as a vector with a magnitude of 2.5m/s and a direction parallel to the sidelines. The receiver, being stationary, has a velocity of 0m/s.

We can also represent the throw as a vector with a magnitude of 8.0m/s and an unknown direction, which we will call θ. Using the concept of vector addition, we can add the velocities of the quarterback and the throw to find the overall velocity of the football. This can be represented by the formula Vf = Vq + Vt, where Vf is the final velocity of the football, Vq is the velocity of the quarterback, and Vt is the velocity of the throw.

Using this formula, we can break down the velocities into their x and y components. The x component of the football's velocity will be 8.0m/scosθ, and the y component will be 8.0m/ssinθ. The x component of the quarterback's velocity will be 2.5m/s, and the y component will be 0m/s.

Since the final velocity of the football must be equal to the velocity of the receiver (0m/s), we can set the x and y components of the final velocity equal to the x and y components of the receiver's velocity. This gives us two equations: 8.0m/scosθ = 0m/s and 8.0m/ssinθ = 0m/s.

Solving for θ, we get θ = 90°, meaning that the throw must be made at a 90° angle relative to the sidelines. To find the distance downfield the receiver is, we can use the formula d = vt, where d is the distance, v is the velocity, and t is the time. Since the receiver is stationary, t = 0. Therefore, the distance downfield is also 0m.

In conclusion, the quarterback must throw the football at a 90° angle relative to the sidelines in order to hit the receiver, who is 0m downfield.