A ball is thrown horizontally from a 15m -high building with a speed of 8.0m/s. How far away from the base of the building does the ball hit the gnd. I'm needing help setting this up. I know change of Y is -15m, and the "a" is -9.8m/s^2, but i get lost from there...?
Using h = Vot + 9.8t^2/2
15 = 0(t) + 4.9t^2 yields t = 1.75 sec to impact.
During this 1.75 sec., the ball travels 1.75(8) = 14m horizontally from the base of the building.
To solve this problem, we can use the motion equations for an object in freefall. Here's how you can set it up:
1. Identify the known values:
- Initial vertical displacement (change in Y) = -15 m (negative because it is downward from the building)
- Initial vertical velocity (Vyi) = 0 m/s (since the ball is thrown horizontally)
- Vertical acceleration (a) = -9.8 m/s^2 (negative because it is directed downward)
2. Use the equation of motion: ΔY = Vyi * t + (1/2) * a * t^2
- Since the initial vertical velocity (Vyi) is 0, the equation simplifies to: ΔY = (1/2) * a * t^2
3. Substitute the known values:
- -15 m = (1/2) * (-9.8 m/s^2) * t^2
4. Solve for time (t):
- Rearrange the equation to isolate t^2: t^2 = (2 * ΔY) / a
- Substitute the values: t^2 = (2 * -15 m) / -9.8 m/s^2
- t^2 = 3.06 s^2
- Taking the square root of both sides: t ≈ 1.75 s (since time cannot be negative)
5. Find the horizontal displacement using the equation: Dx = Vx * t
- Since the ball is thrown horizontally, the horizontal velocity (Vx) remains constant throughout the motion.
6. Determine the horizontal velocity (Vx):
- Use the formula for horizontal velocity: Vx = Initial velocity (8.0 m/s)
7. Substitute the values to find the horizontal displacement (Dx):
- Dx = (8.0 m/s) * 1.75 s
- Dx ≈ 14 m
Therefore, the ball hits the ground approximately 14 meters away from the base of the building.