Part A. With what minimum speed must you toss a 190 g ball straight up to hit the 12 m-high roof of the gymnasium if you release the ball 1.5 m above the ground? Solve this problem using energy. Part B. with what speed does the ball hit the ground?
Intial KE+ initialPE= mg(12)
solve for v in the KE formula.
Notice mass divides out.
there is a trick here, it already is up 1.5 meters,so change the right side to
mg (12-1.5). I bet the teacher put that in just to trick everyone.
To solve this problem using energy, we can apply the principle of conservation of mechanical energy. We consider the initial and final states of the ball and equate the initial mechanical energy with the final mechanical energy.
Part A:
1. First, let's define the initial and final positions of the ball:
Initial position (1): The ball is at a height of 1.5 m above the ground.
Final position (2): The ball reaches a height of 12 m on the roof.
2. The initial mechanical energy of the ball is given by the sum of its kinetic energy (KE) and gravitational potential energy (PE):
Initial mechanical energy (1): E₁ = KE₁ + PE₁
3. At the highest point, when the ball reaches the roof, its kinetic energy becomes zero since it momentarily stops. Therefore, the final mechanical energy is just the gravitational potential energy:
Final mechanical energy (2): E₂ = PE₂
Since mechanical energy is conserved, we have: E₁ = E₂
4. Now, let's calculate the initial mechanical energy (E₁):
KE₁ = (1/2)mv₁², where m = mass of the ball and v₁ = initial velocity of the ball.
PE₁ = mgh, where g = acceleration due to gravity (9.8 m/s²) and h = height above the ground.
E₁ = (1/2)mv₁² + mgh
5. Next, we calculate the final mechanical energy (E₂):
PE₂ = mgh, where h = 12 m (height of the roof).
E₂ = mgh
6. Setting E₁ = E₂, we get:
(1/2)mv₁² + mgh = mgh
7. The masses (m) cancel out, and we solve for the initial velocity (v₁):
(1/2)v₁² = gh
v₁² = 2gh
v₁ = √(2gh)
Substituting the given values, where g = 9.8 m/s² and h = 12 m, we can calculate the minimum speed (v₁) required to hit the roof.
Part B:
To find the speed at which the ball hits the ground, we can use the concept of conservation of energy again.
1. Let's define the initial and final positions of the ball:
Initial position (1): The ball is at a height of 12 m on the roof.
Final position (2): The ball is at a height of 0 m on the ground.
2. Using the same principle as before, we equate the initial mechanical energy with the final mechanical energy:
Initial mechanical energy (1): E₁ = PE₁
Final mechanical energy (2): E₂ = KE₂ + PE₂
3. Since mechanical energy is conserved, we have: E₁ = E₂
4. Now, let's calculate the initial mechanical energy (E₁):
PE₁ = mgh, where g = 9.8 m/s² and h = 12 m.
5. Finally, we calculate the final mechanical energy (E₂):
KE₂ = (1/2)mv₂², where m = mass of the ball and v₂ = final velocity of the ball.
E₂ = KE₂ + PE₂ = (1/2)mv₂² + mgh
Equating E₁ and E₂, we can solve for the final velocity (v₂) to find the speed at which the ball hits the ground.
Remember to substitute the given values and solve the equations to obtain the specific values of the minimum speed required to hit the roof (Part A) and the speed at which the ball hits the ground (Part B) based on the problem's parameters.