For the first part of this question I got 12.12m/s
B. With the help of a friend the boy measures the height to which the ball returns and finds it to be 5.2 meters. How fast was the ball moving when it left the ground?
I wrote this down
mass of ball: 0.40kg
height if house: 7.5
gravity: 9.8m/s^2
v1: 12.12m/s
Help please
v=sqrt(2gh)
To solve this problem, we will use the conservation of mechanical energy. The mechanical energy is conserved when only conservative forces are acting. In this case, the only conservative force is the gravitational force, so we can use the principle of conservation of mechanical energy.
The initial mechanical energy (Ei) of the ball is equal to the final mechanical energy (Ef).
The initial mechanical energy (Ei) is given by the sum of the potential energy (PE) and the kinetic energy (KE) of the ball when it leaves the ground.
The final mechanical energy (Ef) is given by the sum of the potential energy (PE) and kinetic energy (KE) of the ball when it returns to the same height.
Let's calculate the initial mechanical energy Ei first:
Ei = PEi + KEi
The potential energy (PEi) at the initial height is given by:
PEi = mgh
Where m is the mass of the ball (0.40 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the initial height of the ball (7.5 m). Thus,
PEi = (0.40 kg)(9.8 m/s^2)(7.5 m)
Next, we have the initial kinetic energy (KEi) which is given by:
KEi = (1/2)mv1^2
Where m is the mass of the ball (0.40 kg) and v1 is the initial velocity of the ball. We have already found v1 to be 12.12 m/s.
Now, let's calculate the final mechanical energy Ef:
Ef = PEf + KEf
The potential energy (PEf) at the final height is given by:
PEf = mgh
Where m is the mass of the ball (0.40 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the final height of the ball (5.2 m). Thus,
PEf = (0.40 kg)(9.8 m/s^2)(5.2 m)
Next, we have the final kinetic energy (KEf) which is given by:
KEf = (1/2)mv2^2
Where m is the mass of the ball (0.40 kg) and v2 is the final velocity of the ball.
Now, we can set up the equation for conservation of energy:
Ei = Ef
PEi + KEi = PEf + KEf
Plugging in the known values:
(mgh) + [(1/2)mv1^2] = (mgh) + [(1/2)mv2^2]
Canceling out the common terms, we get:
[(1/2)mv1^2] = [(1/2)mv2^2]
v1^2 = v2^2
Rearranging the equation to solve for v2, we get:
v2 = sqrt(v1^2)
Now, we can substitute the known value of v1 into the equation to find v2:
v2 = sqrt((12.12 m/s)^2)
Evaluating the expression, we find:
v2 ≈ 12.12 m/s
So, the speed of the ball when it left the ground is also approximately 12.12 m/s.