How Should the Quarterback Throw to Reach the Moving Receiver?

[bmchenry]

Homework Statement

2 football players are 16m apart. The receiver runs perpendicular to the line joining the two palyers at a constant speed of 7 m/s. after 2 seconds the quarterbak throws the ball with a horizontal velocity of 24m/s

In what horizontal direction and with what vertical launch should the quarterback throw the ball so it reaches the same spot that the receiver will be and what time will the ball be caught?

Homework Equations

once i get the horizontal direction I know how to determine the launch angle and time, I am not sure what to do with the first part

The Attempt at a Solution

I am having trouble setting up the triangle,

 

[AFG34]

so there are three ppl total with the quarterback being one of the 2 players who are 16m apart?

 

[Moetasim]

because I cannot determine the angle between the two players and the receiver's path. However, assuming the receiver continues on a straight path, the quarterback must throw the ball at an angle that is perpendicular to the receiver's path in order for it to reach the same spot at the same time. This means that the ball's launch angle should be 90 degrees.

To determine the time when the ball will be caught, we can use the equation d = vt, where d is the distance between the two players, v is the velocity of the receiver (7 m/s), and t is the time it takes for the receiver to reach the same spot as the ball. Plugging in the values, we get 16m = (7 m/s)t. Solving for t, we get t = 2.29 seconds.

To find the vertical launch of the ball, we can use the equation h = ut + 1/2at^2, where h is the height of the ball, u is the initial velocity (in this case, 24 m/s), a is the acceleration due to gravity (9.8 m/s^2), and t is the time it takes for the ball to reach the same spot as the receiver. Plugging in the values, we get h = (24 m/s)(2.29 s) + 1/2(9.8 m/s^2)(2.29 s)^2. Solving for h, we get h = 27.8 m.

Therefore, the quarterback should throw the ball at a 90 degree angle with a vertical launch of 27.8 meters and it will be caught by the receiver after 2.29 seconds.