a 145g baseball is thrown up at 40 m/s. after traveling a distance of 50m in the air, the baseball is caught. the ball experienced air resistance of what magnitude?
See my answer to the same question of "anonymous" this evening
To find the magnitude of the air resistance experienced by the baseball, we first need to calculate the initial velocity of the baseball using the given information.
Given:
Mass of the baseball (m) = 145g = 0.145 kg
Total distance traveled by the baseball (s) = 50m
Initial velocity of the baseball (u) = 40 m/s
Final velocity of the baseball (v) = 0 m/s (since it's caught)
Acceleration due to gravity (g) = 9.8 m/s^2 (assuming no significant change in height)
We can use the kinematic equation:
v^2 = u^2 + 2as
Rearranging the equation, we can solve for the acceleration (a):
a = (v^2 - u^2) / (2s)
Substituting the known values:
a = (0^2 - 40^2) / (2 * 50)
a = (-1600) / 100 = -16 m/s^2
The negative sign indicates that the acceleration is directed opposite to the initial velocity, which is the deceleration caused by air resistance.
Now, we can calculate the magnitude of the air resistance force (F) using Newton's second law:
F = m * a
Substituting the values:
F = 0.145 kg * (-16 m/s^2)
F = -2.32 N
The magnitude of the air resistance experienced by the baseball is approximately 2.32 Newtons in the opposite direction of its motion. Note that the negative sign indicates the opposite direction as the initial velocity.