91Ó°ÊÓ

For this part, we want to find the highest point the stone can reach. At the highest point, the stone will have a zero vertical velocity. We will use one of the kinematic equations: $$v^2 = u^2 + 2as$$ where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration (negative since it's opposing the motion), and \(s\) is the vertical displacement. We'll rearrange the equation to solve for \(s\): $$s = \frac{v^2 - u^2}{2a}$$ Plugging in our values: \(v = 0\), \(u = 19.6 \,m/s\), \(a = -9.80 \,m/s^2\): $$s = \frac{0^2 - (19.6)^2}{2(-9.80)}$$ After solving for \(s\), we get: $$s = 19.6 m$$ Since the stone was initially \(1.5\,m\) above the ground, the maximum height above the ground will be: Max Height = Initial Height + Vertical Displacement $$Max Height = 1.5m + 19.6m = 21.1m$$

For this part, we need to find the time it takes for the stone to fall back to the ground after reaching its highest point. We'll use the following kinematic equation: $$s = u(t) + \frac{1}{2}a(t)^2$$ Since in our case, \(u = 0\) at the highest point; this simplifies the equation to: $$s = \frac{1}{2}a(t)^2$$ We need to find the time for the stone to hit the ground: $$t = \sqrt{\frac{2s}{a}}$$ As calculated in part (a), the maximum height is \(21.1m\), and \(a = -9.80 \,m/s^2\). We have: $$t = \sqrt{\frac{2(21.1)}{-9.80}}$$ After solving for \(t\), we get: $$t = 2.07s$$ This is the time taken for the stone to descend from its highest point back to the ground. However, we need to find the total time before the stone hits the ground considering its upward motion as well. To do this, we can use the equation: $$v = u + at$$ We'll rearrange the equation to solve for \(t\): $$t = \frac{v - u}{a}$$ In this part, \(u = 19.6\, m/s\), \(v = 0\), and \(a = -9.80\, m/s^2\). So, $$t = \frac{0 - 19.6}{-9.80}$$ After solving for \(t\), we get: $$t = 2s$$ Now, the total time taken for the stone to hit the ground can be found by adding the time for upward motion and the time for downward motion: $$Total Time = 2s + 2.07s = 4.07s$$