A boy throws a stone straight upward with an initial speed of 12.0
Part A
What maximum height will the stone reach before falling back down?
To find the maximum height the stone will reach before falling back down, we need to use the kinematic equation for vertical motion:
vf^2 = vi^2 + 2 * a * d
Where:
- vf is the final velocity (which is 0 since the stone is at its maximum height before falling back down).
- vi is the initial velocity (which is 12.0 m/s).
- a is the acceleration due to gravity (which is -9.8 m/s^2).
- d is the displacement or maximum height we want to calculate.
Rearranging the equation to solve for d, we have:
d = (vf^2 - vi^2) / (2 * a)
Substituting the given values into the equation, we get:
d = (0^2 - 12.0^2) / (2 * -9.8)
Simplifying further, we have:
d = (-144) / (-19.6)
d ≈ 7.35 meters
Therefore, the maximum height the stone will reach before falling back down is approximately 7.35 meters.
To find the maximum height the stone will reach before falling back down, we can use the following formula for the vertical motion of an object:
h = (v₀²) / (2g)
where:
h is the maximum height
v₀ is the initial velocity
g is the acceleration due to gravity (approximately 9.8 m/s² on Earth)
In this case, the stone is thrown straight upward, so the initial velocity v₀ is positive (+12.0 m/s). We can substitute these values into the formula to find the answer.
h = (12.0²) / (2 * 9.8)
= 144 / 19.6
≈ 7.35 meters
Therefore, the maximum height the stone will reach before falling back down is approximately 7.35 meters.