Final answer:
Dave's ball, with a mass of 0.07 kg thrown at 28 m/s, reaches a maximum height of approximately 40 meters. This is calculated using the conservation of energy principle, equating the initial kinetic energy to the potential energy at the ball's highest point.
Step-by-step explanation:
To determine how high the ball goes when Dave throws it straight up into the air, we can use the principles of physics and the conservation of energy. Given that the ball has a mass of 0.07 kg and is thrown at a speed of 28 m/s, we ignore the air resistance as per the question's instruction. The maximum height (h) reached by the ball can be calculated using the equation derived from conservation of mechanical energy: the kinetic energy at the point of release equals the potential energy at the highest point.
The kinetic energy (KE) when the ball is thrown is KE = (1/2)mv^2, where m is the mass and v is the initial velocity. The potential energy (PE) at the highest point is PE = mgh, where g is the acceleration due to gravity (approximately 9.81 m/s^2) and h is the height. When the ball reaches its highest point, its velocity is 0 m/s, so its kinetic energy is also 0, and all the initial kinetic energy has been transferred to potential energy. Therefore, (1/2)mv^2 = mgh.
Solving for h gives us h = (1/2) * (v^2 / g). Plugging in the values, we get h = (1/2) * (28^2 / 9.81) which equals to approximately 40 meters. So, the ball thrown by Dave reaches a height of roughly 40 meters.