quest cns-utexas Gdu ntips:/ qu 020 (part 1 of 4) 1.0 points You throw a softball (of mass 300 g) straight up into the air. It reaches its maximum altitude of 10 m and then returns to you. What is the gravitational potential energy of the softball at its highest position? Assume the ball departed from and returned to ground level. Answer in units of J. 021 (part 2 of 4) 1.0 points Assume no energy is lost by the softball while it is in the air. What is the kinetic energy of the softball as soon as it leaves your hand? Answer in units of J. 022 (part 3 of 4) 1.0 points What is the kinetic energy of the softball when it returns to your hand? Answer in units of J. 023 (part 4 of 4) 1.0 points What is the speed of the ball? Answer in units of m/s.
The Correct Answer and Explanation is:
Let’s solve each part step by step:
Given:
- Mass of the softball, m = 300 g = 0.300 kg
- Maximum height, h = 10 m
- Gravitational acceleration, g = 9.8 m/s²
Part 1: Gravitational Potential Energy at the Highest Point
Formula:PE=mghPE = mghPE=mghPE=(0.300 kg)(9.8 m/s2)(10 m)=29.4 JPE = (0.300\ \text{kg})(9.8\ \text{m/s}^2)(10\ \text{m}) = 29.4\ \text{J}PE=(0.300 kg)(9.8 m/s2)(10 m)=29.4 J
Answer: 29.4 J
Part 2: Kinetic Energy as Soon as It Leaves the Hand
Assuming no energy loss, the kinetic energy when the ball leaves the hand is equal to the potential energy at the highest point.
Answer: 29.4 J
Part 3: Kinetic Energy When It Returns to Your Hand
Since energy is conserved and there is no loss, the kinetic energy on return to your hand is also equal to the energy it had when it was launched.
Answer: 29.4 J
Part 4: Speed of the Ball When It Leaves or Returns
Formula:KE=12mv2KE = \frac{1}{2}mv^2KE=21mv2
Solving for speed:v=2KEm=2⋅29.4 J0.300 kg=196=14 m/sv = \sqrt{\frac{2KE}{m}} = \sqrt{\frac{2 \cdot 29.4\ \text{J}}{0.300\ \text{kg}}} = \sqrt{196} = 14\ \text{m/s}v=m2KE=0.300 kg2⋅29.4 J=196=14 m/s
Answer: 14 m/s
Explanation
Gravitational potential energy is the energy an object has due to its position in a gravitational field. It depends on the object’s mass, its height above a reference level, and the acceleration due to gravity. The formula used is potential energy equals mass times gravitational acceleration times height. In this case, the ball weighs 0.300 kilograms and reaches a height of 10 meters. Therefore, the gravitational potential energy at its highest point is 29.4 joules.
Energy conservation tells us that energy in a system is not created or destroyed, only transformed. So, the kinetic energy the ball has when leaving the thrower’s hand must convert entirely into gravitational potential energy at the highest point, assuming no air resistance or other energy losses. Therefore, the ball’s kinetic energy at launch is also 29.4 joules.
As the ball falls back down, this potential energy is transformed back into kinetic energy. Thus, just before the ball is caught, it regains the same amount of kinetic energy it started with: 29.4 joules.
To find the speed of the ball, we use the kinetic energy formula. Solving for velocity, we use the kinetic energy value and the known mass. This gives a result of 14 meters per second. This speed applies both when the ball leaves the thrower’s hand and when it returns, again due to the conservation of energy in the absence of friction or air resistance.
