+11 an acrobat is launched from a cannon at an angle of 60 above the horizontal. The acrobat is caught by a safety net mounted horizontally at the height from which he was initially launched. suppose the acrobat is launched at a speed of 26m/s. How long from when he is launched does he land in the net? [ans: 4.6s]
+14915 How long from when he is launched does he land in the net?
Hello darkestmoose!
\(The\ proportion\ of\ the\ vertical\ take-off\ speed\ is: \\ v_h=26m/sec\cdot sin(60°)\\ Then\ the\ following\ applies:\\ \color{blue}v_h t =\dfrac{g}{2}t^2\ |\ /t\\ v_h=\dfrac{g}{2}t\ |\ \cdot \frac{2}{g}\\ t= \dfrac{2\cdot v_h}{g}=\dfrac{2\cdot26\frac{m}{sec}\cdot sin(60°)}{9.81\frac{m}{sec^2}}\)
\(t=4.59\ sec\)
The acrobat ends up in the network 4.6 seconds after the start.
!
