A 2.00 kg ball is thrown upward at Some unknown angle from the top of a 20.0 M high building If the initial magnitude of the vel
1 answer:
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(a) 0.456 m/s
The maximum speed of the oscillating mass can be found by using the conservation of energy. In fact:
- At the point of maximum displacement, the mechanical energy of the system is just elastic potential energy:
(1)
where
k = 34.9 N/m is the spring constant
A = 5.0 cm = 0.05 m is the amplitude of the oscillation
- At the point of equilibrium, the displacement is zero, so all the mechanical energy of the system is just kinetic energy:
(2)
where
m = 0.42 kg is the mass
vmax is the maximum speed, which is maximum when the mass passes the equilibrium position
Since the mechanical energy is conserved, we can write (1) = (2):
(b) 0.437 m/s
When the spring is compressed by x = 1.5 cm = 0.015 m, the equation for the conservation of energy becomes:
(3)
where the total mechanical energy can be calculated at the point where the displacement is maximum (x = A = 0.05 m):
So, solving (3) for v, we find the speed when x=1.5 cm:
(c) 0.437 m/s
This part of the problem is exactly identical to part b), since the displacement of the mass is still
x = 1.5 cm = 0.015 m
So, the speed when this is the displacement is
(d) 4.4 cm
In this case, we have that the speed of the mass is 1/2 of the maximum value, so:
And by using the conservation of energy again, we can find the corresponding value of the displacement x:
Answer:
20m
6.9s
Explanation:
The vertical velocity of the ball is 20m/s. We can calculate the kinetic energy which gets transferred to potential energy once it gets to the top.
we can subtitute v = 20m/s and g = 10m/s2
So the ball could go 20m high from the child hand, or 120m fro the bottom of the cliff.
The time it takes for the ball to travels to the top is the time it takes for it to decelerate from 20m/s to 0m/s with gravitational deceleration g = 10m/s2
t = v / g = 20 / 10 = 2s
Then the ball will start accelerating down ward with a constant acceleration of g = 10m/s. In order to cover distance d of 120m from the top to the bottom of the cliff
So the total time it takes is 4.9 + 2 = 6.9s