Hi! Question:A ball is tossed from an upper-story window of a building.The ball is given an initial velocity of 7.80m/s at an angle of 21.0 degrees below the horizontal.It strikes the ground 3.00s later.

A.How far horizontally from the base of the building does the ball strik the ground?
B.Find the height from which the ball was thrown.
C.How long does it take the ball to reach a point 10.0m below the level of launching? Thank you so much for your help

A. Multiply the horizontal component of the initial velocity by the time that the ball is in the air.

7.80 x cos 21 x 3.00 = ?

B. Solve the following equation for vertical height above the ground, for the unknown height H.
y = 0 = H -7.80 sin 21 t - (g/2) t^2
using t = 3.0 seconds. g is the acceleration of gravity, 9.8 m/s^2.

C. Use a different vertical height equation, with t as the unknown and the vertical height above the ground being
y = H -10
= H -7.80 sin 21 t - (g/2) t^2
Note that the H cancels out, leaving
-10 = -7.80 sin 21 t - (g/2) t^2
There will be two roots. Take the one that is positive

To solve this problem, we can use the equations of motion for projectile motion. Let's go step by step to find the answers to each part of the question:

A. How far horizontally from the base of the building does the ball strike the ground?

We can use the horizontal distance formula:

horizontal distance = (initial horizontal velocity) * time

The initial horizontal velocity can be found by decomposing the given initial velocity into horizontal and vertical components. The horizontal component is given by 7.80 m/s * cos(21°).

horizontal velocity = 7.80 m/s * cos(21°) ≈ 7.170 m/s

Now, we can calculate the horizontal distance:

horizontal distance = 7.170 m/s * 3.00 s ≈ 21.51 m

So, the ball strikes the ground approximately 21.51 meters horizontally from the base of the building.

B. Find the height from which the ball was thrown.

To find the height from which the ball was thrown, we can use the vertical distance formula:

vertical distance = (initial vertical velocity) * time + (acceleration due to gravity * time^2)/2

The initial vertical velocity can be found by decomposing the given initial velocity into horizontal and vertical components. The vertical component is given by 7.80 m/s * sin(21°).

vertical velocity = 7.80 m/s * sin(21°) ≈ 2.675 m/s

Using the given values:

vertical distance = (2.675 m/s * 3.00 s) + (9.8 m/s^2 * (3.00 s)^2)/2
= (8.025 m) + (44.1 m)/2
= 8.025 m + 22.05 m
≈ 30.075 m

Therefore, the ball was thrown from a height of approximately 30.075 meters.

C. How long does it take the ball to reach a point 10.0 m below the level of launching?

To find the time it takes for the ball to reach a point 10.0 m below the level of launching, we will need to use the vertical distance formula again.

Vertical distance = (initial vertical velocity) * time + (acceleration due to gravity * time^2)/2

We can rearrange this equation to solve for time:

(time^2 * acceleration due to gravity) / 2 + (initial vertical velocity) * time - vertical distance = 0

Putting the given values into the equation:

(4.9 m/s^2 * time^2) / 2 + (2.675 m/s) * time - 10.0 m = 0

We can solve this quadratic equation using the quadratic formula or factoring. The quadratic formula gives us:

time = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 4.9 m/s^2 / 2 = 2.45 m/s^2, b = 2.675 m/s, and c = -10.0 m.

Plugging these values into the quadratic formula:

time = (-2.675 ± √((2.675)^2 - 4 * 2.45 * (-10.0))) / (2 * 2.45)

Calculating this, we get two solutions:

time ≈ 0.474 s (approximately)
time ≈ 4.434 s (approximately)

The time taken for the ball to reach a point 10.0 m below the level of launching is approximately 0.474 s.

I hope this helps! Let me know if you have any further questions.

Hi! I'd be happy to help you with those questions.

To solve these problems, we can use the equations of motion for a projectile in two dimensions. The horizontal and vertical motion can be treated independently.

First, we need to break down the initial velocity into its horizontal and vertical components.

The horizontal component (Vx) is given by Vx = v * cos(θ), where v is the initial velocity and θ is the angle below the horizontal. Substituting the given values, Vx = 7.80 m/s * cos(21.0°).

The vertical component (Vy) is given by Vy = v * sin(θ). Substituting the given values, Vy = 7.80 m/s * sin(21.0°).

Now, let's solve each question:

A. How far horizontally from the base of the building does the ball strike the ground?

We can use the equation of motion for horizontal displacement: Δx = Vx * t, where t is the time taken for the ball to strike the ground.

Substituting the given values, Δx = (7.80 m/s * cos(21.0°)) * 3.00 s. Calculate this to find the horizontal displacement from the base of the building where the ball strikes the ground.

B. Find the height from which the ball was thrown.

We can use the equation of motion for vertical displacement: Δy = Vy * t + (1/2) * g * t^2, where g is the acceleration due to gravity (approximately 9.8 m/s^2).

Substituting the given values, Δy = (7.80 m/s * sin(21.0°)) * 3.00 s + (1/2) * 9.8 m/s^2 * (3.00 s)^2. Calculate this to 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?

We can use the equation of motion for vertical displacement: Δy = Vy * t + (1/2) * g * t^2. Rearrange the equation to solve for time (t).

Substituting the given values, -10.0 m = (7.80 m/s * sin(21.0°)) * t + (1/2) * 9.8 m/s^2 * t^2. Solve this quadratic equation to find the time it takes for the ball to reach a point 10.0 m below the level of launching.

I hope this helps you solve the problem! Let me know if you have any further questions.