A) With what minimum speed must you toss a 190 ball straight up to hit the 15.0 -high roof of the gymnasium if you release the ball 1.70 above the ground? Solve this problem using energy.
B) With what speed does the ball hit the ground?
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A) To find the minimum speed required to toss the ball, we can use the principle of conservation of energy. The initial energy of the ball is equal to the energy at its maximum height.
The initial energy of the ball consists of its kinetic energy and potential energy. The kinetic energy (KE) is given by 1/2 * m * v^2, where m is the mass of the ball and v is its velocity. The potential energy (PE) is given by m * g * h, where g is the acceleration due to gravity and h is the maximum height.
At the highest point, the velocity of the ball becomes zero, so the kinetic energy is zero. The potential energy at the highest point is equal to the gravitational potential energy at the release point.
Therefore, we can set the initial energy equal to the potential energy at the highest point:
KE + PE = PE
1/2 * m * v^2 + m * g * h = m * g * H
Where H is the maximum height reached (15.0 m) and h is the initial height (1.70 m).
Simplifying the equation:
1/2 * v^2 + g * h = g * H
Now we can solve for the minimum velocity v:
v^2 = 2 * g * (H - h)
v = √(2 * g * (H - h))
Plugging in the given values:
v = √(2 * 9.8 m/s^2 * (15.0 m - 1.70 m))
v = √(2 * 9.8 m/s^2 * 13.30 m)
v ≈ 15.29 m/s
Therefore, the minimum speed required to toss the ball straight up to hit the roof is approximately 15.29 m/s.
B) To find the speed with which the ball hits the ground, we can use the principle of conservation of energy again. The initial energy of the ball at the highest point (where the velocity is zero) will be equal to the final energy of the ball just before it hits the ground.
The final energy of the ball consists of its potential energy at the release point and the kinetic energy just before hitting the ground.
Therefore, we can set the initial energy equal to the final energy:
KE + PE = PE + KE
1/2 * m * v^2 + m * g * h = m * g * H + 1/2 * m * v_g^2
Where v_g is the velocity of the ball just before hitting the ground.
At the highest point, the velocity is zero, so the first term on the left side of the equation becomes zero.
Simplifying the equation:
m * g * h = m * g * H + 1/2 * m * v_g^2
Now we can solve for the velocity v_g:
v_g^2 = 2 * g * (h - H)
v_g = √(2 * g * (h - H))
Plugging in the given values:
v_g = √(2 * 9.8 m/s^2 * (1.70 m - 15.0 m))
v_g = √(2 * 9.8 m/s^2 * (-13.30 m))
v_g ≈ 10.22 m/s
Therefore, the speed with which the ball hits the ground is approximately 10.22 m/s.