Question

exercise hints: getting started | im stuck! a projectile is launched straight up at 58 m/s from a height of 81.5 m, at the edge of a sheer cliff. the projectile falls, just missing the cliff and hitting the ground below. (a) find the maximum height of the projectile above the point of firing. m (b) find the time it takes to hit the ground at the base of the cliff. s (c) find its velocity at impact. m/s

Explanation:

Step1: Find maximum height above firing point

We use the kinematic - equation $v^{2}=v_{0}^{2}-2gh$. At the maximum height, $v = 0$. The initial vertical velocity $v_{0}=58\ m/s$ and the acceleration due to gravity $g = 9.8\ m/s^{2}$.
$0 = 58^{2}-2\times9.8\times h_{max}$
$h_{max}=\frac{58^{2}}{2\times9.8}=\frac{3364}{19.6}\approx171.63\ m$

Step2: Find time to hit the ground

We use the kinematic - equation $y = y_{0}+v_{0}t-\frac{1}{2}gt^{2}$, where $y = 0$ (ground - level), $y_{0}=81.5\ m$, $v_{0}=58\ m/s$, and $g = 9.8\ m/s^{2}$. So, $0 = 81.5+58t-4.9t^{2}$.
Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$. Here, $a=-4.9$, $b = 58$, and $c = 81.5$.
$t=\frac{-58\pm\sqrt{58^{2}-4\times(-4.9)\times81.5}}{2\times(-4.9)}=\frac{-58\pm\sqrt{3364 + 1597.4}}{-9.8}=\frac{-58\pm\sqrt{4961.4}}{-9.8}=\frac{-58\pm70.44}{-9.8}$
We take the positive root $t=\frac{-58 + 70.44}{-9.8}=\frac{12.44}{-9.8}\approx - 1.27$ (rejected) or $t=\frac{-58-70.44}{-9.8}=\frac{-128.44}{-9.8}\approx13.11\ s$

Step3: Find velocity at impact

We use the kinematic - equation $v = v_{0}-gt$. Substituting $v_{0}=58\ m/s$, $g = 9.8\ m/s^{2}$, and $t = 13.11\ s$.
$v=58-9.8\times13.11=58 - 128.48=-70.48\ m/s$ (the negative sign indicates the direction is downward)

Answer:

(a) $171.63$
(b) $13.11$
(c) $-70.48$