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Question
A football quarterback is moving straight backward at a speed of 2.00 m/s when he throws a pass to a player 18.0 m straight downfield. The ball is thrown at an angle of $25.0^\circ$ relative to the ground and is caught at the same height as it is released. What is the initial velocity of the ball relative to the quarterback?
$17.0 \textrm{ m/s, } 22.2^\circ \textrm{ above horizontal}$
Solution Video

# OpenStax College Physics Solution, Chapter 3, Problem 56 (Problems & Exercises) (8:42)

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Submitted by blueFZ07 on Mon, 10/04/2021 - 14:02

At 7:31 the triangles you draw. Where does the 18.0m sit in relation to the 2 end to end vectors? Does Vgq end at 18 or is it beyond the 18?

Submitted by blueFZ07 on Mon, 10/04/2021 - 19:32

To word my question better, 18m range is from the throw to the catch, right? and the addition of the 2 vectors occurs within that 18 right? So your red vector at 7:43 is that 18 displacement - you just moved the Vgq vector around to the end of Vbg vector to make the triangle with the resultant, right?

So, is this equation: Vbqx = 15.17..ms(cos25) + 2.00ms. Essentially this?: Vbqx = 15.17..ms(cos25) + 2.00ms(cos180)
and this for the y: Vbqy = 15.17..ms(sin25) + 2.00(sin180). ?

Submitted by ShaunDychko on Wed, 10/06/2021 - 08:20

Hi Blue, the 18m range is used in the very beginning of the solution to find $v_o$, the initial speed of the ball with respect to the ground. The 18m was used in the range formula to find $v_0$ at 1:53. The 18m range isn't used any more after that. At 7:43 what you're seeing in red is the resultant velocity, not the displacement. This resultant is the vector sum of the velocity of the ball with respect to ground (which we know by 1:53 in the video since the question tells us the launch angle), plus the velocity of the ground with respect to the quarterback (which is worded slightly confusingly - not quarterback with respect to the ground). We need to figure out the red vector to answer the question of the velocity of the ball with respect to the quarterback.
Hope this helps,
Shaun