a ball is tossed from an upper story window of a building given an initial velocity of 8m/s at an angle of 20 below the horizontal.it strikes the ground 2seconds later. how far horizontaly from the base of the building does it strike the ground? find the height from which the ball was thrown.

See Relate Questions: Tue, 2-7-12, 1:11 AM.

To solve this problem, we can use the principles of projectile motion. We'll break the motion of the ball into its horizontal and vertical components.

First, let's find how far horizontally the ball travels from the base of the building. To do this, we need to find the horizontal component of the initial velocity. We can use the formula:

Vx = V * cos(θ),

where Vx is the horizontal component of the velocity, V is the initial velocity of the ball (8 m/s), and θ is the angle below the horizontal (20 degrees).

Vx = 8 * cos(20°) ≈ 7.595 m/s

Now, we know that the ball travels for 2 seconds. Since the horizontal component of velocity remains constant, we can find the horizontal distance traveled using the formula:

dx = Vx * t,

where dx is the horizontal distance traveled and t is the time (2 seconds).

dx = 7.595 m/s * 2 s ≈ 15.19 m

Therefore, the ball strikes the ground approximately 15.19 meters horizontally from the base of the building.

Now let's find the height from which the ball was thrown. To do this, we need to find the vertical component of the initial velocity. We can use the formula:

Vy = V * sin(θ),

where Vy is the vertical component of the velocity, V is the initial velocity of the ball (8 m/s), and θ is the angle below the horizontal (20 degrees).

Vy = 8 * sin(20°) ≈ 2.742 m/s

Next, we can calculate the maximum height reached by the ball using the formula:

h = (Vy^2) / (2 * g),

where h is the maximum height and g is the acceleration due to gravity (approximately 9.8 m/s^2).

h = (2.742 m/s)^2 / (2 * 9.8 m/s^2) ≈ 0.374 m

Therefore, the ball was thrown from a height of approximately 0.374 meters above the ground.

To summarize:
- The ball strikes the ground approximately 15.19 meters horizontally from the base of the building.
- The ball was thrown from a height of approximately 0.374 meters above the ground.