Find the magnitude of the speed

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SUMMARY

The discussion focuses on calculating the initial speed \( v(0) \) required for a quarterback to throw a football to a receiver running away at a constant speed \( v_r \). The key variables involved are the distance \( D \), the time interval \( t_c \), and the acceleration due to gravity \( g \). The correct approach to find the magnitude of the velocity involves using the equation \( v_0^2 = \left(\frac{g t_c}{2}\right)^2 + \left(\frac{D}{t_c} + v_r\right)^2 \), ensuring that both the x and y components are squared and summed appropriately.

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Homework Statement


A quarterback is set up to throw the football to a receiver who is running with a constant vr directly away from the quarterback and is now a distance D away from the quarterback. The quarterback figures that the ball must be thrown at an angle theta to the horizontal and he estimates that the receiver must catch the ball a time interval tc after it is thrown to avoid having opposition players prevent the receiver from making the catch. In the following you may assume that the ball is thrown and caught at the same height above the level playing field. Assume that the y coordinate of the ball at the instant it is thrown or caught is y = 0 and that the horizontal position of the quaterback is x=0.
Use g for the magnitude of the acceleration due to gravity, and use the pictured inertial coordinate system when solving the problem.
Find the speed v(0) with which the quarterback must throw the ball.
Answer in terms of D, tc, vr, and g.


Homework Equations



v= the square root of the velocity squared


The Attempt at a Solution


I got the following components (according to mastering physics, they are correct):
Viy= gtc^2/2tc
(tc=t catch)
Vix= D+vrtc/tc

To get the magnitude of the velocity I put () around both the x and y components of velocity, added them and took the square root, but mastering physics is saying that is wrong. Any ideas? I also need to find the angle.
 
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So you did this:

[tex]\sqrt{(x) + (y)}[/tex]

In actual fact you should have done this:

[tex]\sqrt{x^2 + y^2}[/tex]
 
ScullyX51 said:

Homework Statement


A quarterback is set up to throw the football to a receiver who is running with a constant vr directly away from the quarterback and is now a distance D away from the quarterback. The quarterback figures that the ball must be thrown at an angle theta to the horizontal and he estimates that the receiver must catch the ball a time interval tc after it is thrown to avoid having opposition players prevent the receiver from making the catch. In the following you may assume that the ball is thrown and caught at the same height above the level playing field. Assume that the y coordinate of the ball at the instant it is thrown or caught is y = 0 and that the horizontal position of the quaterback is x=0.
Use g for the magnitude of the acceleration due to gravity, and use the pictured inertial coordinate system when solving the problem.
Find the speed v(0) with which the quarterback must throw the ball.
Answer in terms of D, tc, vr, and g.

Homework Equations



v= the square root of the velocity squared

The Attempt at a Solution


I got the following components (according to mastering physics, they are correct):
Viy= gtc^2/2tc
(tc=t catch)
Vix= D+vrtc/tc

To get the magnitude of the velocity I put () around both the x and y components of velocity, added them and took the square root, but mastering physics is saying that is wrong. Any ideas? I also need to find the angle.

A couple of things.

First your equation for Vx should be D/Tc +Vr

Vy = gTc/2

Vo2 = (gTc/2)2 + (D/Tc + Vr)2
 

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