Question

Your high school quarterback has thrown a football horizontally down the football field. He needs to increase the time the ball is in the air to give the receiver an opportunity to get in the right position to catch the ball. Which factor will increase the flight time of the football?

A. Increasing the horizontal velocity of the ball

B. Decreasing the horizontal velocity of the ball

C. Increasing the vertical height from which the ball is thrown

D. Decreasing the vertical height from which the ball is thrown


self note *11

Answer 1

C. Increasing the vertical height from which the ball is thrown.

Step-by-step explanation:

In this case, we understand that ball experiments a parabolical motion, which is the combination of horizontal uniform motion and vertical uniform accelerated motion, due to gravity. Equations of motion for the ball are described below:

`x = x_(o)+v_(o,x)\cdot t` (Eq. 1)

`y = y_(o) + v_(o,y)\cdot t +(1)/(2)\cdot g\cdot t^(2)` (Eq. 2)

Where:

`x_(o)`,
`x` - Initial and final horizontal positions, measured in meters.

`y_(o)`,
`y` - Initial and final vertical positions, measured in meters.

`v_(o,x)`,
`v_(o,y)` - Initial horizontal and vertical velocities, measured in meters per second.

`g` - Gravitational acceleration, measured in meters per second.

`t` - Time, measured in seconds.

Now we get (Eq. 2) and solve the expression for the time:

`(1)/(2)\cdot g\cdot t^(2)+v_(o,y)\cdot t + (y_(o)-y) = 0`

`t = \frac{-v_(o,y)\pm\sqrt{v_(o,y)^(2)-2\cdot g \cdot (y_(o)-y)}}{g}`

If
`v_(o,y) > 0`,
`g < 0` and
`y_(o) -y < 0`, then maximum time occurs when:

`t = \frac{-v_(o,y)+ \sqrt{v_(o,y)^(2)-2\cdot g \cdot (y_(o)-y)}}{g}`

And,

`v_(o,y)^(2)-2\cdot g\cdot (y_(o)-y)\geq 0`.

If
`y_(o)` is increased, then
`y_(o) - y` goes closer to zero and
`v_(o,y)^(2)-2\cdot g\cdot (y_(o)-y)` becomes greater and time increased. Hence, we conclude that correct answer is C.