   Chapter 3, Problem 38PE

Chapter
Section
Textbook Problem

# 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. (a) If the ball is thrown at an angle of 25° relative to the ground and is caught at the same height as it is released, what is its initial speed relative to the ground? (b) How long does it take to get to the receiver? (c) What is its maximum height above its point of release?

To determine

(a)

The initial speed of ball relative to the ground.

Explanation

Given info:

Quarterback speed relative to the ground vqg=2.00m/s

Let the initial speed of ball relative to the ground be v0

Horizontal distance or Range R=18.0m

Angle of projectile θ0=25°

Acceleration due to gravity g=9.8m/s2

Formula used:

(i) Time of Flight is given as

T=2v0sinθ0g(i)

(ii) x − component of ball relative to ground

vx=v0cosθ0+vqxor,vx=v0cosθ02.00m/s(ii)

(iii) Horizontal distance or Range for the given projectile is given as

R=vx×T(iii)

Here, all alphabets are in their usual meanings.

Calculation:

Now, from equation (i), (ii) and (iii) we have

Horizontal distance or Range for the given projectile

R=vx×TR=(v0cosθ02.00m/s)×2v0sinθ0gR=( v 0cos θ 0×2 v 0sin θ 02

To determine

(b)

Total time taken by the ball to reach the receiver

To determine

(c)

Maximum height reached by the ball from where it has released

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