You throw a ball with a speed of 30 m/s at an angle of 40.0° above the horizontal directly toward a wall. The wall is 22.0 m from the release point of the ball.

(a) How long does the ball take to reach the wall?

(b) How far above the release point does the ball hit the wall?

(c) What are the horizontal and vertical components of its velocity as it hits the wall?

Vo = 30m/s @ 40deg.

Xo = hor. = 30cos40 = 23m/s.
Yo = ver. = 30sin40 = 19.28m/s.

a. Dh = VT = Hor. dist.
T = Dh / V = 22 / 23 = 0.96s =.
2t = T = 0.96,
t = 0.96 / 2 = 0.48s to reach wall.

b. t(up) = T/2 = 0.96 / 2 = 0.48s.
d(up) = Vo*t + 4.9t^2,
d(up)=19.28*0.48 -4.9(0.48)^2 = 8.125m

c. X = Xo = hor = 30cos40 = 23m/s.
Y^2 = Yo^2 + 2gd,
Y^2 = (19.28)^2 - 19.6*8.125 = 212.47,
Y = ver. = 14.6m/s.

To solve this problem, we can use the equations of motion for projectile motion. Here are the steps to find the answers:

Step 1: Resolve the initial velocity into horizontal and vertical components:
The initial velocity (v) of the ball can be divided into its horizontal (v_x) and vertical (v_y) components using trigonometry. Since the angle is given as 40.0° above the horizontal, we can use the sine and cosine functions to find the components.

v_x = v * cos(angle)
v_y = v * sin(angle)

Given:
v = 30 m/s
angle = 40.0°

Let's calculate:

v_x = 30 * cos(40.0°)
v_x = 30 * 0.766
v_x ≈ 22.98 m/s

v_y = 30 * sin(40.0°)
v_y = 30 * 0.6428
v_y ≈ 19.28 m/s

Step 2: Determine the time taken to reach the wall:
Since the horizontal distance to the wall is given, we can find the time taken using the formula:

time = distance / velocity

Given:
distance = 22.0 m
velocity = v_x

Let's calculate:

time = 22.0 / 22.98
time ≈ 0.956 seconds

(a) The ball takes approximately 0.956 seconds to reach the wall.

Step 3: Calculate the vertical distance of the impact point above the release point:
To find the vertical distance traveled by the ball, we can use the formula of motion in the vertical direction:

vertical_distance = (v_y * time) + (0.5 * acceleration * time^2)

Since the ball is thrown directly upward, the acceleration in the vertical direction is -9.8 m/s^2 (due to gravity).

Given:
v_y = 19.28 m/s
time = 0.956 seconds
acceleration = -9.8 m/s^2

Let's calculate:

vertical_distance = (19.28 * 0.956) + (0.5 * -9.8 * 0.956^2)
vertical_distance ≈ 18.41 m

(b) The ball hits the wall approximately 18.41 m above the release point.

Step 4: Determine the horizontal and vertical components of velocity at the impact point:
At the point of impact, only the horizontal component of the velocity remains constant. The vertical component changes due to gravity.

Given:
velocity in the horizontal direction = v_x
velocity in the vertical direction = v_y - (acceleration * time)

Let's calculate:

velocity in the horizontal direction = 22.98 m/s
velocity in the vertical direction = 19.28 - (-9.8 * 0.956)
= 19.28 + 9.388
≈ 28.67 m/s

(c) The horizontal component of the velocity remains 22.98 m/s, and the vertical component of velocity becomes approximately 28.67 m/s as the ball hits the wall.

To answer these questions, we can use the principles of projectile motion.

(a) To find the time it takes for the ball to reach the wall, we can use the horizontal component of the ball's velocity. The horizontal velocity remains constant throughout the motion. The horizontal velocity can be calculated using the formula:

Vx = V * cos(θ)

Where Vx is the horizontal component of the velocity, V is the initial velocity of the ball, and θ is the launch angle.

In this case, V = 30 m/s and θ = 40.0°. So, to find Vx, we can substitute these values into the formula:

Vx = 30 m/s * cos(40.0°)

Using a calculator, we can find that Vx is approximately 22.98 m/s.

Next, we can use the formula for horizontal distance to find the time it takes for the ball to reach the wall:

Time = Distance / Velocity

The horizontal distance from the release point to the wall is given as 22.0 m and the horizontal velocity (Vx) is 22.98 m/s.

Substituting these values into the formula, we get:

Time = 22.0 m / 22.98 m/s

By dividing these values, we find that the time it takes for the ball to reach the wall is approximately 0.958 seconds.

(b) To find how far above the release point the ball hits the wall (vertical distance), we can use the vertical component of the ball's velocity. The vertical velocity changes as the ball moves due to gravity. The vertical component of the velocity can be calculated using the formula:

Vy = V * sin(θ)

Where Vy is the vertical component of the velocity.

Substituting the values V = 30 m/s and θ = 40.0° into the formula, we get:

Vy = 30 m/s * sin(40.0°)

Using a calculator, we find that Vy is approximately 19.24 m/s.

Now, we can use the time calculated in part (a) and the vertical component of velocity (Vy) to find the vertical distance:

Vertical Distance = Vy * Time - 0.5 * g * Time^2

Where g is the acceleration due to gravity, approximately 9.8 m/s^2.

Substituting the values, we get:

Vertical Distance = 19.24 m/s * 0.958 s - 0.5 * 9.8 m/s^2 * (0.958 s)^2

By substituting and solving this equation, the ball hits the wall approximately 9.61 meters above the release point.

(c) The horizontal and vertical components of velocity as the ball hits the wall remain the same as at the release point. The horizontal component is Vx = 22.98 m/s (found in part a), and the vertical component is Vy = 19.24 m/s (found in part b).