A stone is thrown vertically upward with a speed of 22.0 m/s. How fast is it moving when it reaches a height of 12.0 m?
vf^2=Vu^2+2ad
vf^2= 22^2-2*9.8*12
solve for Vf
248.8
15.77
To find the speed of the stone when it reaches a height of 12.0 m, we can use the principles of projectile motion.
1. First, let's analyze the motion of the stone. Since it is thrown vertically upward, its initial velocity is positive (22.0 m/s) and its final velocity when it reaches the topmost point will be zero. The acceleration due to gravity will act in the opposite direction to the initial velocity.
2. We can use the equation of motion: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
3. In the given situation, at the topmost point, v = 0 m/s and s = 12.0 m. Since the stone is moving against gravity, the acceleration (a) will be negative (-9.8 m/s^2, taking g as the acceleration due to gravity).
4. Rearranging the equation, we get u^2 = v^2 - 2as.
5. Plugging in the known values, we have u^2 = 0 - 2*(-9.8)*(12.0).
6. Solving the equation, we find u^2 = 235.2
7. Taking the square root of both sides, we have u ≈ √235.2 ≈ 15.33 m/s.
Therefore, the speed of the stone when it reaches a height of 12.0 m is approximately 15.33 m/s.