From the window of a building, a ball is tossed from a height y0 above the ground with an initial velocity of 8.20 m/s and angle of 23.0° below the horizontal. It strikes the ground 6.00 s later.
(a) 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.)
(b) With the positive x-direction chosen to be out the window, find the x- and y-components of the initial velocity.
(c) Find the equations for the x- and y- components of the position as functions of time. (Use the following as necessary: y0 and t. Let the variable t be measured in seconds.)
(d) How far horizontally from the base of the building does the ball strike the ground?
To answer these questions, we will use the kinematic equations of motion for projectile motion. These equations describe the motion of an object under the influence of gravity when launched at an angle.
Let's break down the problem step by step:
(a) To find the initial coordinates of the ball, we need to determine the initial height (y-coordinate) above the ground. The problem states that the ball is tossed from a height y0 above the ground. Therefore, the initial y-coordinate is simply y0.
(b) To find the x- and y-components of the initial velocity, we need to use trigonometry. The initial velocity is given as 8.20 m/s at an angle of 23.0° below the horizontal.
The x-component of the initial velocity is given by V0x = V0 * cos(theta), where V0 is the initial velocity and theta is the angle.
Substituting the given values: V0x = 8.20 m/s * cos(23.0°)
The y-component of the initial velocity is given by V0y = V0 * sin(theta), where V0 is the initial velocity and theta is the angle.
Substituting the given values: V0y = 8.20 m/s * sin(23.0°)
(c) To find the equations for the x- and y-components of the position as functions of time, we use the kinematic equations of motion.
For the x-coordinate, the equation is x = V0x * t, where x is the horizontal displacement, V0x is the initial x-component of velocity, and t is time.
For the y-coordinate, the equation is y = y0 + V0y * t - (1/2) * g * t^2, where y is the vertical displacement, y0 is the initial y-coordinate, V0y is the initial y-component of velocity, t is time, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the values we have, we get:
x = (8.20 m/s * cos(23.0°)) * t
y = y0 + (8.20 m/s * sin(23.0°)) * t - (1/2) * (9.8 m/s^2) * t^2
(d) To find how far horizontally from the base of the building the ball strikes the ground, we need to find the total horizontal displacement (x-coordinate) when the ball hits the ground. Since we know the total time taken for the ball to hit the ground (6.00 s), we can substitute this value into the equation for x to find the answer.
x = (8.20 m/s * cos(23.0°)) * 6.00 s
By calculating this expression, we will obtain the final answer for part (d).