A quarterback throws a football toward a receiver from a height of 6 ft . The initial vertical velocity of the ball is . At the same time that the ball is thrown, the receiver raises his hands to a height of 8 ft and jumps up with an initial vertical velocity of .

Projectile motion formula:

time, in seconds, since the ball was thrown
height, in feet, above the ground

Complete the system that models the heights of the ball and the receiver’s hands over time.

COMPLETE
After seconds, the receiver’s hands and the ball will be at the same height.
DONE

Ask by Phillips George. in the United States

Mar 22,2025

Question

A quarterback throws a football toward a receiver from a height of 6 ft . The initial vertical velocity of the ball is . At the same time that the ball is thrown, the receiver raises his hands to a height of 8 ft and jumps up with an initial vertical velocity of .

Projectile motion formula:

time, in seconds, since the ball was thrown
height, in feet, above the ground

Complete the system that models the heights of the ball and the receiver’s hands over time.

COMPLETE
After seconds, the receiver’s hands and the ball will be at the same height.
DONE

Ask by Phillips George. in the United States

Mar 22,2025

UpStudy AI · Accepted Answer

To complete the system that models the heights of the ball and the receiver's hands over time, we need to set the two height equations equal to each other and solve for $$ t $$:

1. The height of the ball is given by:
   $$
   h_b = -16 t^{2} + 14 t + 6
   $$

2. The height of the receiver's hands is given by:
   $$
   h_r = -16 t^{2} + 10
   $$

Now, we set $$ h_b $$ equal to $$ h_r $$:
$$
-16 t^{2} + 14 t + 6 = -16 t^{2} + 10
$$

Next, we can simplify this equation by eliminating the $$ -16 t^{2} $$ terms from both sides:
$$
14 t + 6 = 10
$$

Now, we will solve for $$ t $$:
$$
14 t = 10 - 6
$$
$$
14 t = 4
$$
$$
t = \frac{4}{14} = \frac{2}{7}
$$

Thus, after $$ \frac{2}{7} $$ seconds, the receiver's hands and the ball will be at the same height.

Final answer:
After $$ \frac{2}{7} $$ seconds, the receiver's hands and the ball will be at the same height.
After $$ \frac{2}{7} $$ seconds, the receiver's hands and the ball will be at the same height.

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