A ball is shot from the top of a building with an initial velocity of 16m/s at an angle θ = 43∘ above the horizontal.
Part A
What are the horizontal and vertical components of the initial velocity?
If a nearby building is the same height and 55m away, how far below the top of the building will the ball strike the nearby building?
horizontal=16cos43
vertical=16sin43
To find the horizontal and vertical components of the initial velocity, we can use the following trigonometric formulas:
Horizontal component (Vx) = V * cos(θ)
Vertical component (Vy) = V * sin(θ)
Where V is the magnitude of the initial velocity and θ is the angle above the horizontal.
Given:
Initial velocity (V) = 16 m/s
Angle (θ) = 43°
Substituting the values into the formulas, we get:
Vx = 16 * cos(43°)
Vy = 16 * sin(43°)
Using a calculator, we can evaluate these expressions:
Vx ≈ 11.5308 m/s
Vy ≈ 10.8487 m/s
Therefore, the horizontal component of the initial velocity is approximately 11.5308 m/s, and the vertical component is approximately 10.8487 m/s.
To determine how far below the top of the building the ball strikes the nearby building, we can use the equations of motion to analyze the vertical motion of the ball.
Let's denote the time of flight as t. The time it takes for the projectile to hit the ground can be found using the vertical equation of motion:
Vy = gt - 0.5 * g * t^2
Where g is the acceleration due to gravity (approximately 9.8 m/s^2).
In this case, the initial vertical velocity Vy is 10.8487 m/s, and the equation can be rearranged as:
0.5 * g * t^2 - 10.8487 * t = 0
This is a quadratic equation, and we can solve it using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac))/(2a)
Where a = 0.5 * g, b = -10.8487, and c = 0. Plugging in these values we get:
t = (-(-10.8487) ± sqrt((-10.8487)^2 - 4 * 0.5 * 9.8 * 0))/(2 * 0.5 * 9.8)
Simplifying the equation:
t = (10.8487 ± sqrt(118.0461))/9.8
We can disregard the negative root because it represents an unrealistic solution. Using a calculator, we find:
t ≈ 2.2325 s
Therefore, the time of flight of the projectile is approximately 2.2325 seconds.
To find the distance the ball travels horizontally, we can use the horizontal equation of motion:
x = Vx * t
Substituting the values:
x = 11.5308 m/s * 2.2325 s
x ≈ 25.7511 m
Therefore, the ball will strike the nearby building approximately 25.7511 meters away from the top of the building.