Final answer:
To toss a 100 g ball straight up to touch a 10 m high roof, we use conservation of energy to equate potential and kinetic energy, leading to a minimum required initial velocity of approximately 13.2 m/s.
Step-by-step explanation:
To calculate the minimum speed needed to toss a 100 g ball straight up to just touch the 10-meter-high roof of the gymnasium, we utilize the principle of conservation of energy.
The potential energy at the highest point of the ball's trajectory (the roof) is equal to its kinetic energy when it leaves the hand.
The potential energy at the roof (PE) can be calculated as PE = mgh, where m is the mass of the ball, g is the acceleration due to gravity (9.8 m/s²), and h is the height difference between the release point and the roof.
The kinetic energy (KE) at the moment of release is given by KE = (1/2)mv², where v is the initial velocity.
Setting the potential energy equal to the kinetic energy, we have mgh = (1/2)mv².
The mass cancels out, and we solve for v to find the minimum initial velocity required.
Plugging in the numbers: 9.8 m/s² * (10 m - 1.1 m) = (1/2)v², leading us to v = √(9.8 * 8.9 * 2) m/s.
Calculating this gives us a minimum initial velocity of approximately 13.2 m/s.
Therefore, you must toss the ball straight up with a minimum speed of 13.2 m/s to just touch the 10-meter-high roof.