A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 7.70 m/s at an angle of 23.0° below the horizontal. It strikes the ground 6.00 s later. (Ignore air resistance.)

(a) How far horizontally from the base of the building does the ball strike the ground?

(b) Find the height from which the ball was thrown.

(c) How long does it take the ball to reach a point 10.0 m below the level of launching?

Resolve the velocities into vertical (-v0sin(23°) and horizontal (v0cos(23°)) components.

The height of the building can be found from the vertical component.

The horizontal distance is the product of the time (6 seconds) and the horizontal component.

(c) can be found also from the vertical component:
10 = v0t+(1/2)(-g)t²
Solve for t.

To answer these questions, we can use the equations of motion for projectile motion. In projectile motion, we can separate the horizontal and vertical motion of the ball.

(a) To find the horizontal distance from the base of the building where the ball strikes the ground, we need to find the horizontal component of its velocity.

The initial velocity of the ball is given as 7.70 m/s at an angle of 23.0° below the horizontal. To find the horizontal component of velocity, we can use the equation:

Vx = V * cos(θ)

where Vx is the horizontal component of velocity, V is the initial velocity of the ball, and θ is the angle below the horizontal.

Substituting the given values:

Vx = 7.70 m/s * cos(23.0°)

Calculating Vx:

Vx = 7.70 m/s * 0.9205

Vx ≈ 7.09 m/s

Now, we need to find the time it takes for the ball to hit the ground. We can use the equation of motion:

y = Vyt - (1/2)gt^2

where y is the vertical distance, Vy is the vertical component of velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is time.

Since the ball is launched from an upper-story window, we can assume its initial vertical position is 0. Therefore:

y = -gt^2/2

Substituting values:

0 = -9.8 m/s^2 * (6.00 s)^2 / 2

Calculating:

0 = -9.8 m/s^2 * 18.00 s^2 / 2

0 = -88.2 m * s^2

This equation gives us the vertical distance the ball will fall. Since the ball strikes the ground, the vertical distance is equal to 0. Now we can solve for time:

0 = -88.2 m * s^2 / 2

0 = -44.1 m * s^2

Since the coefficient on the right side is 0, time doesn't matter for the vertical motion of the ball. Therefore, we can conclude that it takes 6.00 seconds for the ball to reach the ground.

Now, we can find the horizontal distance traveled by the ball:

distance = Vx * time

Substituting values:

distance = 7.09 m/s * 6.00 s

Calculating:

distance ≈ 42.54 m

Therefore, the ball strikes the ground approximately 42.54 meters horizontally from the base of the building.

(b) To find the height from which the ball was thrown, we can use the equation of motion for vertical displacement:

y = Vyt - (1/2)gt^2

where y is the vertical distance, Vy is the vertical component of velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is time.

We can solve for y:

y = Vyt - (1/2)gt^2

Substituting values:

y = 7.70 m/s * sin(23.0°) * (6.00 s) - (1/2) * 9.8 m/s^2 * (6.00 s)^2

Calculating:

y ≈ 16.82 m

Therefore, the ball was thrown from a height of approximately 16.82 meters.

(c) To find how long it takes for the ball to reach a point 10.0 m below the level of launching, we can use the equation of motion for vertical displacement:

y = Vyt - (1/2)gt^2

We know the vertical distance, so we can rearrange the equation to solve for time:

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

Substituting values:

t = sqrt((2 * 10.0 m) / 9.8 m/s^2)

Calculating:

t ≈ sqrt(2.04 s^2)

t ≈ 1.43 s

Therefore, it takes approximately 1.43 seconds for the ball to reach a point 10.0 meters below the level of launching.