A quarterback throws a football toward a receiver from a height of 6 ft. The initial vertical velocity of the ball is [tex]$14 ft/s$[/tex]. 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 [tex]$10 ft/s$[/tex].

Projectile motion formula:
[tex]
\begin{array}{l}
h=-16 t^2+v t+h_0 \\
t=\text { time, in seconds, since the ball was thrown } \\
h=\text { height, in feet, above the ground }
\end{array}
[/tex]
Complete the system that models the heights of the ball and the receiver's hands over time.
[tex]
\begin{array}{l}
h=-16 t^2+14 t+6 \\
h=-16 t^2+10 t+8
\end{array}
[/tex]
After [ ] seconds, the receiver's hands and the ball will be at the same height.

The receiver's hands and the ball are at the same height at [ ] ft above the ground.

Set the height equations equal to each other: − 16 t 2 + 14 t + 6 = − 16 t 2 + 10 t + 8 .
Solve for t : 4 t = 2 , which gives t = 0.5 .
Substitute t = 0.5 into either equation to find the height.
The height is h = − 16 ( 0.5 ) 2 + 14 ( 0.5 ) + 6 = 9 ft. The final answer is 9 ​ .

Explanation

Problem Analysis Let's analyze the problem. We are given two equations that represent the height of the ball and the receiver's hands as a function of time. We want to find the time when the ball and the receiver's hands are at the same height, and what that height is.

Setting the Equations Equal To find the time when the ball and the receiver's hands are at the same height, we need to set the two equations equal to each other: − 16 t 2 + 14 t + 6 = − 16 t 2 + 10 t + 8

Solving for Time Now, let's solve for t :
− 16 t 2 + 14 t + 6 = − 16 t 2 + 10 t + 8 14 t + 6 = 10 t + 8 14 t − 10 t = 8 − 6 4 t = 2 t = 4 2 ​ t = 0.5 So, the ball and the receiver's hands are at the same height after 0.5 seconds.

Finding the Height Now that we have the time, we can find the height at that time by substituting t = 0.5 into either equation. Let's use the first equation: h = − 16 t 2 + 14 t + 6 h = − 16 ( 0.5 ) 2 + 14 ( 0.5 ) + 6 h = − 16 ( 0.25 ) + 7 + 6 h = − 4 + 7 + 6 h = 9 So, the ball and the receiver's hands are at the same height at 9 ft above the ground.

Final Answer Therefore, after 0.5 seconds, the receiver's hands and the ball will be at the same height, which is 9 ft above the ground.


Examples
Understanding projectile motion is crucial in sports like basketball or baseball. For example, calculating the trajectory of a basketball shot helps players adjust their angle and velocity for a successful basket. Similarly, in baseball, knowing how a ball will travel when hit allows fielders to position themselves effectively to catch it. These calculations involve factors like initial velocity, launch angle, and gravity, all of which influence the path of the projectile.

Answered by GinnyAnswer | 2025-07-08

After 0.5 seconds, the receiver's hands and the ball will be at the same height of 9 feet above the ground.
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Answered by Anonymous | 2025-07-12