Question

The turd is launched at a speed of 72 m/s at an angle of 12 derees above the horizontal. at exactly what time after launch should the ground be covered in a portal to intercept the turd before it hits the ground?

Answer 1

Answer:

$t \approx 3.053\,s$

Explanation:

The kinematic equations for the turd are described below:

$x = (72\,\frac{m}{s}\cdot \cos 12^{\textdegree} )\cdot t$

$y = (72\,\frac{m}{s}\cdot \sin 12^{\textdegree})\cdot t - \frac{1}{2}\cdot (9.807\,\frac{m}{s^{2}} )\cdot t^{2}$

The maximum horizontal distance coincides with minimum height, then:

$(72\,\frac{m}{s}\cdot \sin 12^{\textdegree})\cdot t - \frac{1}{2}\cdot (9.807\,\frac{m}{s^{2}} )\cdot t^{2}= 0$

Whose roots are:

$t_{1} \approx 3.053\,s$

$t_{2} = 0\,s$

First root is related to the maximum horizontal, whereas the other one has to do with initial position of the turd. The needed time is:

$t \approx 3.053\,s$