a 145g baseball is dropped from a tree 13.0m above the ground. with what is the velocity of the baseball at the ground after it falls 13.0m? B. If it actually hits the ground with a speed of 8.0o m/s, what is the average force of air resistance exerted on it?
Use conservation of energy to calculate the velocity at the ground with no friction.
V = sqrt (2 g H) = 15.97 m/s
If there is a lower velocity of 8.0 m/s, which is about 1/2 of the frictionless value, than about 3/4 of the available potential energy has been lost to friction.
Friction energy loss = (3/4)*M g H = ?
For the average force, solve
(friction energy loss)= (average force) x (13.0 m)
To find the velocity of the baseball at the ground, we can use the equation of motion. Since the initial velocity is zero (the baseball is dropped), the equation simplifies to:
v^2 = u^2 + 2as
Where:
v = final velocity (which we need to find)
u = initial velocity (zero in this case)
a = acceleration due to gravity (-9.8 m/s^2)
s = displacement (13.0 m)
Plugging in the values, the equation becomes:
v^2 = 0 + 2*(-9.8)*13.0
v^2 = -254.8
Since the velocity cannot be negative, we take the positive square root:
v = √254.8
v ≈ 15.97 m/s
Therefore, the velocity of the baseball at the ground is approximately 15.97 m/s.
B. To find the average force of air resistance exerted on the baseball, we can use Newton's second law of motion, which states that force (F) equals mass (m) multiplied by acceleration (a). In this case, the acceleration is the change in velocity (∆v) over the change in time (∆t).
F = m * ∆v/∆t
Since we're given the speed at which the baseball hits the ground (8.0 m/s), we can find the change in velocity:
∆v = v_final - v_initial
∆v = 8.0 - 0
∆v = 8.0 m/s
The time taken to travel the 13.0 meters can be found using the equation of motion:
s = ut + (1/2)at^2
Rearranging the equation, we get:
t = sqrt(2s/a)
t = sqrt(2 * 13.0 / 9.8)
t ≈ 1.85 seconds
Now, we can calculate the average force of air resistance using Newton's second law:
F = m * ∆v/∆t
F = 0.145 kg * (8.0 m/s) / (1.85 s)
F ≈ 0.627 N
Therefore, the average force of air resistance exerted on the baseball is approximately 0.627 Newtons.