From the window of a building, a ball is tossed from a height y 0 above the ground with an initial velocity of 10.00 m/s and angle of 25.0° below the horizontal. It strikes the ground 3.60 s later.
If the base of the building is taken to be the origin of the coordinates, with upward the positive y- direction, determine:
a) How far horizontally from the base of the building does the ball strike the ground (5 points)
b) the height from which the ball was thrown (5 points)
s) the velocity with which it hits the ground below (5 points)
Vo = 10m/s[-25o]
Xo = 10*cos(-25) = 9.063 m/s
Yo = 10*sin(-25) = -4.226 m/s.
a. D=Xo * Tf = 9.063m/s * 3.60s=32.63 m.
b. h = Yo*t + 0.5g*t^2
h = -4.226*3.6 + 4.9*3.6^2 = 48.29 m.
c. Y^2 = Yo^2 + 2g*h
Y^2 = -4.226^2 + 19.6*48.29 = 964.3 m.
Y = 31.05 m/s.
To determine the horizontal distance the ball strikes the ground (a), the height (y0) from which the ball was thrown (b), and the velocity with which it hits the ground (s), we can use the equations of motion for projectile motion.
1. Horizontal Distance (a):
In projectile motion, the horizontal component of velocity remains constant throughout the motion. We can use the equation:
a = Vx * t
where "a" is the horizontal distance, "Vx" is the horizontal component of velocity, and "t" is the time.
We can find the horizontal component of velocity using:
Vx = V * cos(theta)
where "V" is the initial velocity and "theta" is the launch angle.
Using the given values, V = 10.00 m/s and theta = -25.0° (since it's below the horizontal), we can calculate Vx as:
Vx = 10.00 m/s * cos(-25.0°)
Now, substitute the values into the equation for horizontal distance:
a = Vx * t
2. Height (b):
The height from which the ball was thrown can be determined using the equation:
y = y0 + V0y * t + (1/2) * g * t^2
where "y" is the final height (ground level), "y0" is the initial height, "V0y" is the vertical component of velocity, "t" is the time, and "g" is the acceleration due to gravity (-9.8 m/s^2).
To find "V0y", the vertical component of velocity, we can use:
V0y = V * sin(theta)
Using the given values, V = 10.00 m/s and theta = -25.0°, we can calculate V0y as:
V0y = 10.00 m/s * sin(-25.0°)
Now, substitute the values into the equation for height:
y = y0 + V0y * t + (1/2) * g * t^2
3. Velocity at Ground (s):
The velocity at which the ball hits the ground can be determined using the equation:
Vf = Vy + g * t
where "Vf" is the final velocity (velocity at the ground), "Vy" is the vertical component of velocity, "g" is the acceleration due to gravity, and "t" is the time.
Using the given values for V0y and t, we can calculate Vf as:
Vf = V0y + g * t
Now, you can substitute the values into each equation to calculate the results for (a), (b), and (s).