Question

1. Ryan throws a baseball upward with an initial velocity of 12 m/s from a height of 2m above the ground. How long will it take for the baseball to hit the ground? Round answer to the nearest 10th of a second.

2. A ball is thrown directly upward from an initial height of 200 feet with an initial velocity of 96 feet per second. How long does it take for the ball to reach its maximum height?

a - 3 seconds

b - 6 seconds

c - 344 seconds

d - 200 seconds


3. A ball is thrown directly upward from an initial height of 200 feet with an initial velocity of 96 feet per second. What is the maximum height that the ball will reach?

a - 3 feet

b - 224 feet

c - 344 feet

d - 200 feet

Answer 1

1 idk

2 B

3 A

hope those 2 help

Answer 2

1.

t = 0.9 sec

2.

Option a is correct. t = 3 sec

3.

Option C is correct, h(3) = 344 feet

Explanation:

1.

The equation that determine the Ryan baseball is given by:

`h(t) = -(1)/(2)at^2+v_0t+h_0` ....[1]

where,

h(t) is the height after t time

a is the acceleration i.e, a ≈ 10 m/s^2

`v_0` is the initial velocity

`h_0` is the initial height from the ground.

As per the statement:

Ryan throws a baseball upward with an initial velocity of 12 m/s from a height of 2m above the ground.

⇒
`v_0 =12 m/s` and
`h_0=2 m`

Substitute the given values we have;

`h(t) = -5t^2+12t+2`

We have to find how long will it take for the baseball to hit the ground.

⇒ h(t) = 0

then;

`0 = -5t^2+12t+2`

Simplify:

t = 2.556 and t = -0.156

Since, t cannot be negative.

Therefore, 2.6 sec long will it take for the baseball to hit the ground.

2.

Given that:

`v_0 = 96 ft/s` and
`h_0 = 200 ft` and use a ≈ 32ft/s^2

then;

`h(t) = -16t^2+96t+200`

To find the maximum height:

A quadratic equation
`y=ax^2+bx+c` then the axis of symmetry is given by:"

`x = -(b)/(2a)`

For
`h(t) = -16t^2+96t+200` we have;

Using axis of symmetry;

`t = -(96)/(2 \cdot -16) = (96)/(32) =3` sec

Therefore, 3 sec does it take for the ball to reach its maximum height.

3.

Similar ques as ques no 2.

We have to find maximum height that the ball will reach

Substitute t =3 sec in
`h(t) = -16t^2+96t+200` to find h(3) we have;

`h(3) = -16(3)^2+96(3)+200 = -144+288+200 = 344 ft`

Therefore, the the maximum height that the ball will reach is, 344 feet