What is the maximum height achieved if a 0.800 kg mass is thrown straight upward with an initial speed of 40.0 m·s–1? Ignore the effect of air resistance.
My physics is quite rusty but I believe that would be
V^2 = Vo^2 + 2ax
V^2 will be zero at the top since eventually the pull of gravity will stop the balls upward flight.
0 = (40 m/s)^2 + 2*(-9.8)*x
Solve for x = distance in meters.
Thats what it looks like.. but im taking chem. in college and have never had a physics course, im really lost.
alright, thanks lol
To find the maximum height achieved by the mass, we need to use the principles of projectile motion. First, let's examine the initial velocity and acceleration of the mass.
Given:
Mass (m) = 0.800 kg
Initial velocity (u) = 40.0 m/s (upward)
Acceleration due to gravity (g) = 9.8 m/s² (downward)
The mass is thrown straight upward, so the acceleration is acting in the opposite direction of the initial velocity. Therefore, the initial acceleration (a) can be calculated as -g.
Next, we can use the kinematic equation that relates the final velocity (v), initial velocity (u), acceleration (a), and displacement (s):
v² = u² + 2as
Since the mass reaches its maximum height, the final velocity at that point becomes 0 m/s. Let's substitute the known values into the equation:
0² = (40.0)² + 2 (-9.8) s
Simplifying the equation gives:
0 = 1600 - 19.6s
Rearranging the equation to isolate 's' (displacement):
19.6s = 1600
s = 1600 / 19.6
Evaluating this expression gives us:
s ≈ 81.63 meters
Therefore, the maximum height achieved by the mass is approximately 81.63 meters.