A) With what minimum speed must you toss a 190 g ball straight up to hit the 15.0 m-high roof of the gymnasium if you release the ball 1.70 m above the ground? Solve this problem using energy.
B) With what speed does the ball hit the ground?
A) To solve this problem using energy, we can equate the potential energy of the ball at the ground level with the potential energy of the ball when it reaches the roof.
The potential energy of the ball at ground level can be calculated using the equation: PE = mgh, where m is the mass of the ball, g is the acceleration due to gravity, and h is the height above the ground.
Given:
Mass of the ball (m) = 190 g = 0.19 kg
Height above the ground (h) = 1.70 m
Height of the roof (H) = 15.0 m
Acceleration due to gravity (g) = 9.8 m/s^2
The potential energy at ground level is:
PE1 = mgh1 = 0.19 kg * 9.8 m/s^2 * 1.70 m
The potential energy at the roof level is:
PE2 = mgh2 = 0.19 kg * 9.8 m/s^2 * (15.0 m + 1.70 m)
Since energy is conserved, we can equate the potential energies at ground level and roof level:
PE1 = PE2
0.19 kg * 9.8 m/s^2 * 1.70 m = 0.19 kg * 9.8 m/s^2 * (15.0 m + 1.70 m)
Simplifying the equation:
1.6334 J = 33.3684 J
We can rearrange the equation to solve for the minimum speed (v) required to reach the roof:
v = sqrt(2gh2)
Plugging in the values:
v = sqrt(2 * 9.8 m/s^2 * (15.0 m + 1.70 m))
Simplifying the equation:
v = sqrt(2 * 9.8 m/s^2 * 16.7 m)
v = sqrt(329.36 m^2/s^2)
v ≈ 18.15 m/s
Therefore, the minimum speed required to toss the ball straight up to hit the 15.0 m-high roof of the gymnasium is approximately 18.15 m/s.
B) To find the speed with which the ball hits the ground, we can use the principle of conservation of energy.
At roof level, the only energy the ball has is kinetic energy, given by the equation: KE = 1/2 mv^2, where m is the mass of the ball and v is its velocity.
Given:
Mass of the ball (m) = 190 g = 0.19 kg
Velocity of the ball at roof level (v) = ?
The kinetic energy at roof level is:
KE = 1/2 mv^2
The total mechanical energy (potential energy + kinetic energy) at roof level is equal to the potential energy at ground level, which we calculated in part A:
PE2 + KE = PE1
0.19 kg * 9.8 m/s^2 * (15.0 m + 1.70 m) + 1/2 * 0.19 kg * v^2 = 0.19 kg * 9.8 m/s^2 * 1.70 m
Simplifying the equation:
33.3684 J + 0.0953 v^2 = 1.6334 J
Rearranging the equation to solve for v:
0.0953 v^2 = 1.6334 J - 33.3684 J
0.0953 v^2 = -31.735 J
v^2 = -31.735 J / 0.0953
v^2 = -333.3878 m^2/s^2
Since velocity cannot be negative, it means that the ball does not hit the ground. This is likely due to the initial velocity not being sufficient to reach the ground.
To solve both parts of the problem (A and B) using energy, we can make use of the conservation of energy principle. According to this principle, the total mechanical energy of an object remains constant as long as no external forces are acting on it.
Let's start with part A:
A) With what minimum speed must you toss a 190 g ball straight up to hit the 15.0 m-high roof of the gymnasium if you release the ball 1.70 m above the ground?
To find the minimum speed required, we can compare the initial mechanical energy (at the release point) with the final mechanical energy (at the highest point, just before hitting the roof).
At the release point:
Initial potential energy = m * g * h (where m is the mass, g is the acceleration due to gravity, and h is the height)
At the highest point:
Final potential energy = m * g * H (where H is the height, which is the distance to the roof)
Since there is no change in kinetic energy (as the ball reaches its highest point), the initial kinetic energy is zero.
Therefore, we can set up the equation:
Initial potential energy = Final potential energy
m * g * h = m * g * H
We can cancel out the mass (m) from both sides of the equation:
g * h = g * H
Now, we can solve for the minimum speed by rearranging the equation:
V = √(2 * g * h)
Substituting the given values:
V = √(2 * 9.8 m/s^2 * 1.7 m)
V ≈ 8.81 m/s
So, the minimum speed required to hit the roof of the gymnasium is approximately 8.81 m/s.
Moving on to part B:
B) With what speed does the ball hit the ground?
To find the speed of the ball when it hits the ground, we can use the conservation of energy principle again. This time, we compare the final mechanical energy at the highest point to the mechanical energy at the ground level.
At the highest point:
Initial potential energy = m * g * H
At ground level:
Final kinetic energy = (1/2) * m * V^2 (where V is the final velocity of the ball)
Again, we set up the equation:
Initial potential energy = Final kinetic energy
m * g * H = (1/2) * m * V^2
We can cancel out the mass (m) from both sides of the equation:
g * H = (1/2) * V^2
Now, we can solve for the speed (V) by rearranging the equation:
V = √(2 * g * H)
Substituting the given values:
V = √(2 * 9.8 m/s^2 * 15.0 m)
V ≈ 19.56 m/s
So, the speed at which the ball hits the ground is approximately 19.56 m/s.