a .17 kg ball of dough is thrown straight up into the air with an initial speed of 12 m/s. The acceleration of gravity is 9.81 m/s^2. What is its momentum halfway to its maximum height on the way up?
To find the momentum halfway to the maximum height, we first need to determine the maximum height the ball of dough can reach when thrown straight up.
The ball of dough will experience deceleration as it moves against the force of gravity. At the maximum height, the ball will momentarily come to a stop before reversing direction.
To find the maximum height (h), we can use the equation of motion:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s at maximum height)
u = initial velocity (12 m/s)
a = acceleration (-9.81 m/s^2, as it acts against the motion)
s = displacement (maximum height, h)
0^2 = 12^2 + 2(-9.81)h
0 = 144 - 19.62h
19.62h = 144
h = 144 / 19.62
h ≈ 7.35 meters (maximum height)
Now that we know the maximum height is 7.35 meters, we can find the halfway point, which is 7.35 / 2 = 3.675 meters.
The momentum of an object is given by the formula:
p = m * v
where:
p = momentum
m = mass (0.17 kg)
v = velocity
To find the velocity at the halfway point, we can use the equation of motion:
v = u + at
where:
v = final velocity
u = initial velocity (12 m/s)
a = acceleration (-9.81 m/s^2)
t = time
At the halfway point, the final velocity will be zero. We need to find the time taken to reach this velocity.
0 = 12 - 9.81t
9.81t = 12
t = 12 / 9.81
t ≈ 1.224 seconds
Now we can find the velocity at the halfway point:
v = 12 - 9.81 * 1.224
v ≈ 12 - 12
v ≈ 0 m/s
Therefore, the velocity at the halfway point is approximately 0 m/s.
Now, we can calculate the momentum at the halfway point:
p = m * v
p = 0.17 kg * 0 m/s
p = 0 kg∙m/s
Thus, the momentum halfway to its maximum height on the way up is 0 kg∙m/s.