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?

**Problem 3** A stone is dropped (initial velocity \( u = 0 \)) and falls under gravity until its velocity reaches \( v = 250 \, \mathrm{m/s} \). 1. **Time to reach \(250 \, \mathrm{m/s}\):** Using the relation \[ v = u + gt, \] where \( g = 9.8 \, \mathrm{m/s^2} \), we set \[ 250 = 0 + 9.8t. \] Solving for \( t \): \[ t = \frac{250}{9.8} \approx 25.51 \, \mathrm{s}. \] 2. **Distance fallen:** Using the displacement formula \[ s = ut + \frac{1}{2}gt^2, \] and noting \( u = 0 \), we have \[ s = \frac{1}{2} \cdot 9.8 \cdot t^2 = 4.9t^2. \] Substitute \( t \approx 25.51 \, \mathrm{s} \): \[ s \approx 4.9 \times (25.51)^2 \approx 4.9 \times 650.76 \approx 3188.72 \, \mathrm{m}. \] --- **Problem 4** A ball is thrown upward with an initial speed \( u = 20 \, \mathrm{m/s} \) from the top of a building of height \( h_0 = 12 \, \mathrm{m} \). **(a) Height of the ball above the ground at its highest point** 1. **Displacement from the point of throw to the highest point:** At the highest point, the final velocity is \( v = 0 \). Using \[ v^2 = u^2 - 2gs, \] we have \[ 0 = (20)^2 - 2(9.8)s. \] Solving for \( s \): \[ s = \frac{20^2}{2 \times 9.8} = \frac{400}{19.6} \approx 20.41 \, \mathrm{m}. \] 2. **Total height above the ground:** Add the building’s height: \[ H = h_0 + s \approx 12 + 20.41 \approx 32.41 \, \mathrm{m}. \] **(b) Time to reach the highest point** Using \[ v = u - gt, \] with \( v = 0 \): \[ 0 = 20 - 9.8t. \] Solving for \( t \): \[ t = \frac{20}{9.8} \approx 2.04 \, \mathrm{s}. \] **(c) Total time taken for the ball to reach the ground** We use the vertical motion equation with the ground as \( y = 0 \) and set upward as positive. The initial conditions are: - Initial height: \( y_0 = 12 \, \mathrm{m} \) - Initial velocity: \( u = 20 \, \mathrm{m/s} \) - Acceleration: \( a = -9.8 \, \mathrm{m/s^2} \) The displacement equation is \[ y = y_0 + ut - \frac{1}{2}gt^2. \] Setting \( y = 0 \): \[ 0 = 12 + 20t - 4.9t^2. \] This can be rearranged to: \[ 4.9t^2 - 20t - 12 = 0. \] Using the quadratic formula \[ t = \frac{20 \pm \sqrt{20^2 - 4(4.9)(-12)}}{2 \cdot 4.9}, \] compute the discriminant: \[ \Delta = 20^2 - 4(4.9)(-12) = 400 + 235.2 = 635.2. \] Thus, \[ t = \frac{20 \pm \sqrt{635.2}}{9.8}. \] Since time must be positive, take the positive root. Approximating: \[ \sqrt{635.2} \approx 25.20, \] so \[ t = \frac{20 + 25.20}{9.8} \approx \frac{45.20}{9.8} \approx 4.61 \, \mathrm{s}. \] --- **Final Answers** - **Problem 3:** Time to reach \(250 \, \mathrm{m/s}\) is approximately \(25.51 \, \mathrm{s}\); distance fallen is approximately \(3188.72 \, \mathrm{m}\). - **Problem 4:** (a) Height above ground at the highest point: approximately \(32.41 \, \mathrm{m}\). (b) Time to reach the highest point: approximately \(2.04 \, \mathrm{s}\). (c) Total time to reach the ground: approximately \(4.61 \, \mathrm{s}\).