From the window of a building, a ball is tossed from a height y0 above the ground with an initial velocity of 8.50 m/s and angle of 23.0° below the horizontal. It strikes the ground 6.00 s later. If the base of the building is taken to be the origin of the coordinates, with upward the positive y-direction, what are the initial coordinates of the ball? (Use the following as necessary: y0.)
Vo = 8.50m/s[-23o]
Xo = 8.50*Cos(-23) = 7.82 m/s.
Yo = 8.50*sin(-23) = -3.32 m/s. = Ver.
component of initial velocity.
To find the initial coordinates of the ball, we need to determine the initial height and the horizontal distance it traveled before reaching the ground.
Given:
- The initial velocity of the ball is 8.50 m/s.
- The angle of the ball's initial trajectory below the horizontal is 23.0°.
- The time it takes for the ball to strike the ground is 6.00 s.
First, let's calculate the initial vertical velocity (Vy0) and the initial horizontal velocity (Vx0) of the ball.
Vy0 is given by the equation:
Vy0 = V0 * sin(θ)
Vx0 is given by the equation:
Vx0 = V0 * cos(θ)
Where:
- V0 is the initial velocity of the ball (8.50 m/s).
- θ is the angle of the ball's initial trajectory below the horizontal (23.0°).
Plugging in the values, we get:
Vy0 = 8.50 m/s * sin(23.0°)
Vy0 ≈ 3.712 m/s
Vx0 = 8.50 m/s * cos(23.0°)
Vx0 ≈ 7.733 m/s
Next, let's calculate the initial height (y0) and the horizontal distance traveled (x) by the ball.
The equation for the vertical displacement (y) of an object in free fall is given by:
y = y0 + Vy0 * t + (1/2) * g * t^2
Since the ball starts at a height y0 above the ground, the initial vertical displacement is 0, and the equation becomes:
0 = y0 + Vy0 * t + (1/2) * g * t^2
Solving for y0:
y0 = -Vy0 * t - (1/2) * g * t^2
Where:
- g is the acceleration due to gravity (approximately 9.8 m/s^2).
- t is the time it takes for the ball to strike the ground (6.00 s).
Plugging in the values, we get:
y0 = -3.712 m/s * 6.00 s - (1/2) * 9.8 m/s^2 * (6.00 s)^2
y0 ≈ -133.43 m
The negative sign indicates that the initial height (y0) is below the origin (ground level).
Now, let's calculate the horizontal distance traveled (x) by the ball.
The equation for the horizontal displacement (x) of an object in uniform motion is given by:
x = Vx0 * t
Plugging in the values, we get:
x = 7.733 m/s * 6.00 s
x ≈ 46.40 m
Therefore, the initial coordinates of the ball are (-46.40 m, -133.43 m).