Question

A ball is thrown vertically in the air with a velocity of 95 ft/s. The ball is at a height of 120 ft.

Answer 1

To determine the ball's initial velocity, we can use the equation for free fall motion. Since the ball is thrown vertically, the initial vertical velocity is the same as the final vertical velocity when the ball reaches its maximum height.

Step-by-step explanation:

The subject of this question is Physics and it is a question for High School level.

To determine the ball's initial velocity, we can use the equation for free fall motion. Since the ball is thrown vertically, the initial vertical velocity is the same as the final vertical velocity when the ball reaches its maximum height. The initial vertical velocity can be determined using the height and time taken for the ball to pass the window.

The initial velocity of the ball can be found using the formula:

Vertical component of velocity (Vy) = Final velocity (Vfy)

Since the final velocity is 0 when the ball reaches its maximum height, we can substitute the height and time values to find the initial vertical velocity. Once we have the initial vertical velocity, we can assume no horizontal acceleration and use the formula:

Velocity (V) = Velocity in x-direction (Vx)

Since the ball is thrown vertically, the horizontal component of velocity remains constant throughout the motion. Therefore, the initial horizontal velocity is the same as the final horizontal velocity.

Answer 2

The ball is at a height of 120 feet after 1.8 and 4.1 seconds.

Step-by-step explanation:

The statement is incomplete. The complete statement is perhaps the following "A ball is thrown vertically in the air with a velocity of 95 ft/s. What time in seconds is the ball at a height of 120ft. Round to the nearest tenth of a second."

Since the ball is launched upwards, gravity decelerates it up to rest and moves downwards. The position of the ball can be determined as a function of time by using this expression:

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

Where:

`y_(o)` - Initial height of the ball, measured in feet.

`v_(o)` - Initial speed of the ball, measured in feet per second.

`g` - Gravitational constant, equal to
`-32.174\,(ft)/(s^(2))`.

`t` - Time, measured in seconds.

Given that
`y_(o) = 0\,ft`,
`v_(o) = 95\,(ft)/(s)`,
`g = -32.174\,(ft)/(s^(2))` and
`y = 120\,ft`, the following second-order polynomial is found:

`120\,ft = 0\,ft + \left(95\,(ft)/(s) \right)\cdot t +(1)/(2)\cdot \left(-32.174\,(ft)/(s^(2)) \right) \cdot t^(2)`

`-16.087\cdot t^(2) + 95\cdot t -120 =0`

The roots of this polynomial are, respectively:

`t_(1) \approx 4.075\,s` and
`t_(2) \approx 1.831\,s`.

Both roots solutions are physically reasonable, since
`t_(1)` represents the instant when the ball reaches a height of 120 ft before reaching maximum height, whereas
`t_(2)` represents the instant when the ball the same height after reaching maximum height.

In nutshell, the ball is at a height of 120 feet after 1.8 and 4.1 seconds.