3. A stone is dropped from the fop of a mountain. Assuming no air resistance, how far will it fall and for how long in order to reach a velocity of \( 250 \mathrm{~m} .- \) ? 4. A ball is thrown upward at \( 20 \mathrm{~m} \cdot \mathrm{~s}^{-1} \) off the top of a building that is 12 m high. a) What is the height of the ball above the ground at its highest point? b) How long does it take the ball to reach its highest point? c) What is the total time taken for the ball to reach the ground after leaving the person's hand?

When a stone is dropped from the top of a mountain, it accelerates due to gravity at approximately \(9.81 \, \text{m/s}^2\). To find how far it will fall to reach a velocity of \(250 \, \text{m/s}\), you can use the formula \(v^2 = u^2 + 2as\), where \(v\) is the final velocity, \(u\) is the initial velocity (which is 0 when dropped), \(a\) is the acceleration, and \(s\) is the distance fallen. Plugging in the values gives \(250^2 = 0 + 2(9.81)s\), solving for \(s\) results in approximately \(3,186 \, \text{m}\) or \(3.2 \, \text{km}\). To find the time it takes to reach this velocity, you can use \(v = u + at\). Rearranging gives \(t = \frac{v - u}{a} = \frac{250 - 0}{9.81} \approx 25.5 \, \text{s}\). So, it takes about 25.5 seconds to reach that velocity while falling through approximately 3.2 kilometers. For the ball thrown upward, at its highest point, the velocity will be zero. To find its maximum height, use the kinetic energy formula: \(h = \frac{v^2}{2g} + h_i\), where \(h_i\) is the initial height of the building (12 m). Substituting the values gives \(h = \frac{(20)^2}{2 \times 9.81} + 12 \approx 20.4 \, \text{m}\). The ball reaches a height of roughly 20.4 meters above the ground. To determine how long it takes to reach this highest point, use \(t = \frac{v-u}{g}\). Here, the final velocity is 0 (at the peak), the initial velocity is \(20 \, \text{m/s}\), and \(g\) is \(9.81 \, \text{m/s^2}\). So, \(t \approx \frac{0 - 20}{-9.81} \approx 2.04 \, \text{s}\). Finally, for the total time taken for the ball to reach the ground after being thrown, you need to find the time taken to reach the highest point (2.04 s) and then falling from that max height of 20.4 m to the ground. Use the equation \(s = ut + \frac{1}{2}gt^2\) set to \(h\), where \(u = 0\) during the fall. Solving this will give you the falling time, adding it to the ascent time to find total time. Overall, it will take about 4.58 seconds in total for the ball to reach the ground.