Chapter 9.5, Problem 75E

To determine

To calculate: The minimum initial vertical velocity $v_0$ needed to reach the branch.

Answer to Problem 75E

The required minimum initial velocity is 24.66 feet per second.

Explanation of Solution

Given information:

The height of the tree $=15$ feet

Height of the weight at which it is released $=5.5$ feet

The given height of the tree that is, $h=15$ feet and height of the weight that is, $s_0=5.5$ feet.

To find the minimum initial vertical velocity

  $v_0$ , use the equation $h=-16t^2+v_{0}t+s_0$ .

Substitute 15 for h , 5.5 for $s_0$ in the above equation.

  $\begin{array}{c}-16t^2+v_{0}t+5.5=15\\ 16t^2-v_{0}t-5.5+15=0\\ 16t^2-v_{0}t+9.5=0\end{array}$

To solve the above equation, use the discriminant rule.

The discriminant of the above quadratic equation has one real solution; if the discriminant is zero.

  $\begin{array}{c}{b}^2-4ac=0\\ {\left(-v_0\right)}^2-4\left(16\right)\left(9.5\right)=0\\ v_0{}^2-608=0\\ v_0{}^2=608\end{array}$

Take square root on each side of the above equation.

  $\begin{array}{c}\sqrt{v_0{}^2}=\sqrt{608}\\ v_0\approx \pm 24.66\end{array}$

But, the velocity cannot be negative.

So, $v_0\approx 24.66$

Hence, the required minimum initial velocity is 24.66 feet per second.