Question

7.) A screw was detached from a hot air balloon that is rising at 3.5 m/s at an altitude of 8.3 m. (a) What is the initial velocity of the screw? (b) How long will it take for the screw to reach the ground?

Answer 1

First a reference system is established.

Where semi positive axis points up + and

Question (a)

The initial velocity of the screw is equal to the initial velocity of the balloon at the time the screw separates, therefore

$\boxed{V_{oy}=3.5\frac{m}{s}}$ .

Question (b)

The screw position is given by:

$y_{f}=y_{o}+V_{oy}*t+\frac{1}{2}a_{y}t^{2}$

Where.

• Final position (when it reaches the ground)

$y_{o}=8.5m$

• Initial position

$y_{f}=0m$

• Initial velocity

$V_{0y}=3.5\frac{m}{s}$

• Gravity acceleration

$a=-9.8\frac{m}{s^{2}}$

Numerically evaluating and organizing

 

A second grade equation.

$0=8.5+3.5t-\frac{9.8}{2}t^{2}$

$-4.9t^{2}+3.5t+8.5=0-->ax^{2}+bx+c=0$

Coefficients $a=-4.9;b=3.5;c=8..5$

solving numerically

$t=\frac{-b\pm \sqrt{b^{2}-4*a*c}}{2*a}$

$t=\frac{-(3.5)\pm \sqrt{(3.5)^{2}-4*-4.9*8.5}}{2*-4.9}$

• Solution 1

$t=\frac{-(3.5)+ \sqrt{(3.5)^{2}-4*-4.9*8.5}}{2*-4.9}=-1.00s$

• Solution 2

$t=\frac{-(3.5)- \sqrt{(3.5)^{2}-4*-4.9*8.5}}{2*-4.9}=1.72s$

The time considered is positive.

The time it takes to get to the ground is:

$\boxed{t=1.72s}$