Question

With what minimum speed must you toss a 100g ball straight up to just touch the 15-m-high roof of the gymnasium if you release the ball 1.5m above the ground? Solve this problem using energy.

Part B
With what speed does the ball hit the ground?

Answer 1

By using the conservation of energy principle, the minimum speed required to touch the gymnasium roof is approximately 16.26 m/s. The speed at which the ball hits the ground after falling 1.5m is approximately 17.15 m/s.

Step-by-step explanation:

Calculating the Minimum Launch Speed and Impact Speed

To find the minimum speed to touch the 15-m-high roof, we use the conservation of energy principle. The kinetic energy at the moment of release must equal the gravitational potential energy at the top of the roof. The initial kinetic energy (KE) can be expressed as KE = 0.5 * m * v2, where m is the mass of the ball and v is the velocity. The potential energy (PE) at the roof is given by PE = m * g * h, where g is the acceleration due to gravity (9.8 m/s2) and h is the height difference (15m - 1.5m = 13.5m).

Setting these equal, we find the minimum speed, v, required:

0.5 * m * v2 = m * g * h

• v = sqrt(2 * g * h)

Plugging in the values: v = sqrt(2 * 9.8 * 13.5), we find that the initial velocity is approximately 16.26 m/s.

For the second part, when the ball returns to the initial height, its speed will be the same as the launch speed due to conservation of energy in a vacuum. However, it will be faster when it hits the ground because it falls an additional 1.5 meters. The final speed can be calculated using:

• PE at 1.5m + KE at 1.5m = KE at ground level
• m * g * 1.5 + 0.5 * m * v2 = 0.5 * m * v'2
• v' = sqrt(v2 + 2 * g * 1.5)

Substituting the given values, we find that the speed at which the ball hits the ground is approximately 17.15 m/s.

Answer 2

Using conservation of mechanical energy, the minimum speed required to toss a 100g ball to touch a 15-m-high gymnasium roof is found by equating the potential energy at the roof height to the kinetic energy at the release point. The speed at which the ball hits the ground will be the same as the initial speed, due to energy conservation.

Step-by-step explanation:

To solve for the minimum speed required to toss a 100g ball to just touch the 15-m-high gymnasium roof, we will use the principle of conservation of mechanical energy. We consider the point of release at 1.5m as the reference point, where the potential energy (PE) will be 0. The total energy needed for the ball to reach the height just below the roof, 13.5m from the release point, is all potential energy at the top.

Using the formula for gravitational potential energy PE = mgh, where m is the mass, g is the gravity (9.8 m/s2), and h is the height, we find the PE at the roof. Substituting m = 0.1 kg, g = 9.8 m/s2, and h = 13.5 m, we get PE = (0.1 kg)(9.8 m/s2)(13.5 m).

The ball must have this same amount of kinetic energy (KE) at the release point to reach the roof. The KE is given by KE = 0.5mv2 where v is the velocity we want to find. Equating KE to the calculated PE, we can solve for v.

For part B, the ball will hit the ground with the same speed it had when it was released because energy is conserved unless the ball loses energy during its flight, which we ignore as per the instructions. Thus, the speed at which the ball hits the ground is equal to the initial speed found in part A.