+72 An acrobat is launched from a cannon at an angle of 60o 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 26 m/s.
+118608 An acrobat is launched from a cannon at an angle of 60o 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 26 m/s.
How long does it take before he reaches his maximum height?
How long does it take in total for him to reach a point halfway back down to the ground?
\(\text{Vertical components}\\ \dot{y}(0)=26sin60=22.517m/s\\ \ddot{y}=-9.8\\ \dot{y}=-9.8t+22.517\\ y=-4.9t^2+22.517t\\ \text{When y=0}\\ -4.9t^2+22.517t=0\\ t(-4.9t+22.517)=0\\ t=0\;\;or\;\;t=\frac{22.514}{4.9}=4.595\;\;sec\\ \text{To get to the max height will be reached in half this time, approx 2.3 secs }\\ When\; t=2.3sec,\;\;y \approx 25.87m \)
Half way down the y value is
0.5*25.87 = 12.935m
When y=12.935 find t
\(-4.9t^2+22.517t=12.935\\ 4.9t^2-22.517t+12.935=0\\ \)
4.9t^2-22.517t+12.935=0 = {t=-(((sqrt(253489289)-22517)/9800)), t=((sqrt(253489289)+22517)/9800)}
t=0.6730, t=3.9222
It takes about 3.9 secs for the acrobat to be half way down :)
You should check all this as it is easy to make careless errors 
