Question

A football quarterback throws a pass at an angle of 41.3°. He releases the pass 3.50 m behind the line of scrimmage. His receiver left the line of scrimmage 2.3 s earlier, going straight down-field at a constant speed of 7.50 m/s. With what speed must the quarterback throw the ball so that the pass lands gently in the receiver's hands without the receiver breaking stride? Assume that the ball is released at the same height it is caught and that the receiver is straight downfield from the quarterback at the time of release. (Ignore any effects due to air resistance.)

Answer 1

Answer on Question #70264-Physics-Other

A football quarterback throws a pass at an angle of 41.3°. He releases the pass 3.50 m behind the line of scrimmage. His receiver left the line of scrimmage 2.3 s earlier, going straight down-field at a constant speed of 7.50 m/s. With what speed must the quarterback throw the ball so that the pass lands gently in the receiver's hands without the receiver breaking stride? Assume that the ball is released at the same height it is caught and that the receiver is straight downfield from the quarterback at the time of release. (Ignore any effects due to air resistance.)

Solution

The horizontal distance of projectile is

$$x = v \cos \alpha t$$

The horizontal distance of the receiver is

$$x' = 3.50 + 7.50 (t + 2.3)$$

The time of flight of projectile is

$$t = \frac{2 v \sin \theta}{g}$$ $$x = x'$$ $$v \cos \alpha \frac{2 v \sin \theta}{g} = 3.50 + 7.50 \left(\frac{2 v \sin \theta}{g} + 2.3\right)$$ $$\frac{v^2 \sin 2\theta}{9.81} = 3.50 + 7.50 \left(\frac{2 v \sin \theta}{g} + 2.3\right)$$ $$\frac{v^2 \sin 2(41.3{}^\circ)}{9.81} = 3.50 + 7.50 \left(\frac{2 v \sin 41.3{}^\circ}{9.81} + 2.3\right)$$ $$v_1 = 20.2 \frac{m}{s}.$$ $$v_2 = -10.2 \frac{m}{s}.$$

The second answer is impossible (negative magnitude of velocity).

Answer: $20.2 \frac{m}{s}$.

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