A suspension bridge is \(60.0 \mathrm{m}\) above the level base of a gorge. A stone is thrown or dropped from the bridge. Ignore air resistance. At the location of the bridge \(g\) has been measured to be $9.83 \mathrm{m} / \mathrm{s}^{2} .$ (a) If you drop the stone, how long does it take for it to fall to the base of the gorge? (b) If you throw the stone straight down with a speed of \(20.0 \mathrm{m} / \mathrm{s},\) how long before it hits the ground? (c) If you throw the stone with a velocity of \(20.0 \mathrm{m} / \mathrm{s}\) at \(30.0^{\circ}\) above the horizontal, how far from the point directly below the bridge will it hit the level ground?

In this case, the stone is simply dropped from the bridge. We can use the following equation of motion to find the time it takes for the stone to fall to the base of the gorge: \(h = \frac{1}{2}gt^2\) where \(h = 60.0\) m, \(g=9.83\ \mathrm{m/s^2}\), and \(t\) is the time it takes for the stone to fall. Solving for \(t\): \(t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2(60.0)}{9.83}} \approx 3.49 \ \mathrm{s}\) So, the time it takes for the stone to fall is \(3.49 \ \mathrm{s}\).

In this case, the stone is thrown straight down with a speed of \(20.0 \mathrm{m/s}\). Using the following equation of motion: \(h = v_0t + \frac{1}{2}gt^2\) where \(h = 60.0\) m, \(v_0= 20.0 \ \mathrm{m/s}\), \(g=9.83\ \mathrm{m/s^2}\), and \(t\) is the time it takes for the stone to fall. Solving for \(t\): \(t= \frac{-v_0 + \sqrt{v_0^2+2gh}}{g} = \frac{-20 +\sqrt{(20)^2+2(9.83)(60)}}{9.83} \approx 2.42\ \mathrm{s}\) So, when the stone is thrown straight down with a speed of \(20.0 \mathrm{m/s}\), it takes \(2.42\ \mathrm{s}\) for it to hit the ground.

In this case, the stone is thrown with a velocity of \(20.0 \mathrm{m/s}\) at \(30.0^{\circ}\) above the horizontal. First, we need to find the vertical and horizontal components of the velocity: \(v_0x = v_0 \cos(30^\circ) = 20\ \mathrm{m/s} \times \cos(30^\circ) \approx 17.32\ \mathrm{m/s}\) \(v_0y = v_0 \sin(30^\circ) = 20\ \mathrm{m/s} \times \sin(30^\circ) = 10.0\ \mathrm{m/s}\) Now, we'll find the time it takes for the stone to hit the ground using the equation of motion: \(h = v_0yt + \frac{1}{2}gt^2\) Solving for \(t\): \(t= \frac{-v_0y + \sqrt{v_0y^2+2gh}}{g} = \frac{-10 +\sqrt{(10)^2+2(9.83)(60)}}{9.83} \approx 2.90\ \mathrm{s}\) Finally, we'll find the distance from the point directly below the bridge that the stone hits the ground using the following equation: \(x = v_0xt\) \(x = (17.32\ \mathrm{m/s})(2.90\ \mathrm{s}) \approx 50.23\ \mathrm{m}\) So, when the stone is thrown with a velocity of \(20.0\ \mathrm{m/s}\) at \(30.0^\circ\) above the horizontal, it will hit the ground approximately \(50.23\ \mathrm{m}\) from the point directly below the bridge.