HELP :)
A man who is standing on the roof of a building throws a baseball vertically downward toward the sidewalk at a speed of 6.85m/s. The sidewalk is 34.9m below the roof. Calculate the speed of the ball just as it hits the pavement.
h = Vo*t + 0.5g*t^2 = 34.9.
6.85t + 4.9t^2 = 34.9,
4.9t^2 + 6.85t - 34.9 = 0,
Quad. Formula:
t = (-6.85 +- sqrt(6.85^2 + 684.04))/9.8 = 2.06 s. to hit the pavement.
V = Vo + g*t = 6.85 + 9.8*2.06 = 20.2 m/s.
Correction: V = 27.1 m/s.
Another Method:
V^2 = Vo^2 + 2g*d = 6.85^2 + 19.6*34.9 = 730.963.
V = 27.05 m/s.
To solve this problem, we can use the equations of motion and relevant physics principles.
First, let's identify the given values:
Initial velocity (vi) = 6.85 m/s (thrown vertically downward)
Height (h) = 34.9 m (vertical distance from the roof to the sidewalk)
Acceleration due to gravity (g) = 9.8 m/s² (assumed and constant on Earth's surface)
Now, let's use the following equation of motion to find the final velocity (vf) of the ball just as it hits the pavement:
vf² = vi² + 2gh
Here's the explanation of the equation:
- vf is the final velocity of the ball
- vi is the initial velocity of the ball
- g is the acceleration due to gravity
- h is the vertical distance traveled by the ball
Substituting the known values into the equation:
vf² = (6.85 m/s)² + 2 * (9.8 m/s²) * (34.9 m)
Now, calculate the right-hand side of the equation:
vf² = 46.9225 m²/s² + 683.24 m²/s²
Add the values:
vf² = 730.1625 m²/s²
Finally, take the square root of both sides to find the final velocity, vf:
vf ≈ √(730.1625 m²/s²)
vf ≈ 27.02 m/s
Therefore, the speed of the ball just as it hits the pavement is approximately 27.02 m/s.