A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 8.15 m/s at an angle of 18.5o below the horizontal. It strikes the ground 3.30 s later. How far horizontally from the base of the building does the ball strike the ground?
What heigh was the ball thrown from?
Vo = 8.15 m/s @ 18.5 Deg.,down.
Xo = 8.15*cos18.5 = 7.73 m/s.
Yo = 8.15*sin18.5 = 2.59 m/s.
a. Tf = 3.30 s. = Fall time or time in
flight.
Dx=Xo * Tf = 7.73m/s * 3.30s = 25.5 m.=
Hor. dist. from base of bldg.
b. h = Yo*t + 0.5g*t^2,
h = 2.59*3.30 + 4.9*(3.3)^2 = 61.9 m.
To find the horizontal distance the ball strikes from the base of the building, we can use the horizontal component of the initial velocity and the time it takes for the ball to strike the ground.
The horizontal component of the velocity can be found using the formula:
Vx = V * cos(θ)
Where:
Vx is the horizontal component of the velocity,
V is the initial velocity (8.15 m/s),
and θ is the angle below the horizontal (18.5 degrees).
Plugging in the known values:
Vx = 8.15 m/s * cos(18.5)
Next, we can use the horizontal velocity and the time to calculate the horizontal distance traveled by the ball. We'll use the formula:
d = Vx * t
Where:
d is the horizontal distance,
Vx is the horizontal component of the velocity (from the previous step),
and t is the time taken for the ball to reach the ground (3.30 seconds).
Plugging in the known values:
d = Vx * t
Now, let's calculate the values:
Vx = 8.15 m/s * cos(18.5)
≈ 7.79 m/s
d = 7.79 m/s * 3.30 s
≈ 25.68 meters
So, the ball strikes the ground approximately 25.68 meters horizontally from the base of the building.
To find the height from which the ball was thrown, we need to calculate the vertical displacement (or height) using the vertical component of the initial velocity, the time, and the acceleration due to gravity.
The vertical component of the velocity can be found using the formula:
Vy = V * sin(θ)
Where:
Vy is the vertical component of the velocity,
V is the initial velocity (8.15 m/s),
and θ is the angle below the horizontal (18.5 degrees).
Plugging in the known values:
Vy = 8.15 m/s * sin(18.5)
Next, we can use the equation of motion to calculate the vertical displacement. The equation is:
y = Vy * t + (1/2) * a * t^2
Where:
y is the vertical displacement,
Vy is the vertical component of the velocity (from the previous step),
t is the time taken for the ball to reach the ground (3.30 seconds),
and a is the acceleration due to gravity (-9.8 m/s^2).
Plugging in the known values:
y = Vy * t + (1/2) * a * t^2
Now, let's calculate the values:
Vy = 8.15 m/s * sin(18.5)
≈ 2.74 m/s
y = 2.74 m/s * 3.30 s + (1/2) * (-9.8 m/s^2) * (3.30 s)^2
≈ 4.58 meters
So, the ball was thrown from a height of approximately 4.58 meters.
To find the horizontal distance the ball strikes the ground, we need to calculate the projectile's horizontal displacement. We can use the equation:
Horizontal displacement = initial velocity * time
Given that the initial velocity of the ball is 8.15 m/s and the time of flight is 3.30 s, we can substitute these values into the equation to get the horizontal displacement:
Horizontal displacement = 8.15 m/s * 3.30 s
Horizontal displacement = 26.895 m
Therefore, the ball strikes the ground approximately 26.895 meters horizontally from the base of the building.
To find the height from which the ball was thrown, we can calculate the vertical displacement of the ball using the equation:
Vertical displacement = initial velocity * sin(angle) * time + (1/2) * acceleration * time^2
In this equation, the initial velocity is 8.15 m/s, the angle is 18.5 degrees, the time is 3.30 s, and the acceleration due to gravity is approximately -9.8 m/s^2 (taking downward direction as negative).
Vertical displacement = 8.15 m/s * sin(18.5°) * 3.30 s + (1/2) * (-9.8 m/s^2) * (3.30 s)^2
Vertical displacement = 11.356 m - 51.147 m
Vertical displacement ≈ -39.791 m
The negative sign indicates that the ball was thrown downward from the building. Hence, the height from which the ball was thrown is approximately 39.791 meters below the upper-story window.