A baseball hit just above the ground leaves the bat at 27 m/s at 45∘ above the horizontal.

A)How far away does the ball strike the ground?

B)How much time is the ball in the air?

C)What is its maximum height?

Find the time for the ball to go the horizontal distance to home plate. Then find how far the ball drops in this time.

u = 27 cos 45 forever

Vi = 27 sin 45 at start

h = Vi t -4.9 t^2 = t (Vi -4.9t)
h = 0 at ground
so at ground at t = 0 of course and at
t = 27 sin 45 /4.9

range = u t

max height at t = half that t
h = Vi t - 4.9 t^2

second question nonsense, from where

To solve this problem, let's break it down step-by-step:

Step 1: Find the initial vertical and horizontal velocities of the ball.
Given:
- Initial velocity (v0) = 27 m/s
- Launch angle (θ) = 45°

To find the initial vertical velocity (v0y) and initial horizontal velocity (v0x), we can use the following equations:

v0y = v0 * sin(θ)
v0x = v0 * cos(θ)

Substituting the given values:
v0y = 27 m/s * sin(45°)
v0x = 27 m/s * cos(45°)

Step 2: Calculate the time the ball takes to reach the horizontal distance to home plate.
The horizontal distance to home plate is unknown, so let's call it "d."
We can use the equation for horizontal distance as follows:

d = v0x * t

Since we want to find the time (t), let's rearrange the equation:

t = d / v0x

Step 3: Find how far the ball drops in this time.
To determine how far the ball drops vertically, we need to find the vertical displacement (Δy).
Since acceleration due to gravity (g) acts only vertically, we can use the following equation:

Δy = v0y * t + (1/2) * g * t^2

Substituting the values:

Δy = (v0 * sin(θ)) * t + (1/2) * g * t^2

Now, let's solve the problem step-by-step:

A) How far away does the ball strike the ground?
To calculate the horizontal distance, we need the time it takes, which we can find using the equation:

t = d / v0x

Since the question doesn't provide a specific horizontal distance, we cannot solve part A without additional information.

B) How much time is the ball in the air?
We can calculate the time using the equation:

t = d / v0x

Since the question doesn't provide a specific horizontal distance, we cannot solve part B without additional information.

C) What is its maximum height?
The maximum height occurs when the ball's vertical velocity becomes zero. To find it, we can use the equation:

v0y - g * t = 0

Substituting the values:

(27 m/s * sin(45°)) - (9.8 m/s^2 * t) = 0

Solving for t:

t = (27 m/s * sin(45°)) / (9.8 m/s^2)

Then, we substitute the value of t into the equation for vertical displacement:

Δy = (27 m/s * sin(45°)) * [(27 m/s * sin(45°)) / (9.8 m/s^2)] + (1/2) * 9.8 m/s^2 * [(27 m/s * sin(45°)) / (9.8 m/s^2)]^2

Simplifying further, substitute sin(45°) = √2 / 2:

Δy = (27 m/s * √2 / 2) * [(27 m/s * √2 / 2) / (9.8 m/s^2)] + (1/2) * 9.8 m/s^2 * [(27 m/s * √2 / 2) / (9.8 m/s^2)]^2

Evaluate these calculations to find the maximum height of the ball.

To find the distance the ball strikes the ground (A), we can first calculate the time it takes for the ball to hit the ground. We'll need to consider the vertical motion of the ball:

1. Resolve the initial velocity into its vertical and horizontal components:

Vertical component: V_y = V * sin(45°)
Horizontal component: V_x = V * cos(45°)

where V is the initial velocity of the ball (27 m/s) and sin/cos are trigonometric functions.

2. Use the vertical motion equation to find the time it takes for the ball to hit the ground:

h = (1/2) * g * t^2

where h is the initial vertical displacement (0 as it is hit just above the ground), g is the acceleration due to gravity (typically 9.8 m/s^2), and t is the time.

Since the initial vertical displacement is 0, the equation simplifies to:

0 = (1/2) * g * t^2

Solving for t, we get:

t = sqrt((2 * h) / g)

where sqrt is the square root function.

3. Substitute the vertical motion equation into the horizontal motion equation to find the horizontal distance traveled:

d = V_x * t

where d is the horizontal distance traveled.

Now, let's proceed with solving each question step by step:

A) How far away does the ball strike the ground?

1. Resolve the initial velocity:
V_y = 27 m/s * sin(45°)
V_x = 27 m/s * cos(45°)

2. Calculate the time:
t = sqrt((2 * h) / g) (since initial vertical displacement is 0)

3. Find the horizontal distance traveled:
d = V_x * t

B) How much time is the ball in the air?

The time it takes for the ball to hit the ground (computed in step 1) is the same as the time the ball is in the air.

C) What is its maximum height?

The maximum height is achieved when the vertical velocity component becomes 0 due to the effect of gravity. We can calculate this time using the following equation:

V_y = V * sin(45°) - g * t_max = 0

Solving for t_max, we get:

t_max = (V * sin(45°)) / g

Then, we can substitute this time into the vertical motion equation (h = (1/2) * g * t^2) to find the maximum height.