+12 Write a quadratic equation that has the projectile reach a maximum height of 45 meters
+14995 a maximum height of 45 meters.
Hello SBAHarper!
The launch velocity is \(v_0.\)
The launch angle to the plane is \(\alpha.\)
Let the running time be t.
\({\color{blue}g}=9.805m/sec^2\)
\(h=-\frac{g}{2}\cdot t^2+v_0\cdot sin\ \alpha \cdot t\\ \dfrac{dh}{dt}=-g\cdot t+v_0\cdot sin\ \alpha=0\\ t=\dfrac{v_0\cdot sin\ \alpha}{g}\)
\(h_{max}=45m\\ =-\frac{g}{2}(\dfrac{v_0\cdot sin\ \alpha}{g})^2+v_0\cdot sin\ \alpha \cdot \dfrac{v_0\cdot sin\ \alpha}{g}\\ =-\dfrac{v_0^2\cdot sin^2\alpha}{2g}+\dfrac{2v_0^2\cdot sin^2\alpha}{2g}\)
\(h_{max}(=45m)=\dfrac{3}{2}\cdot \dfrac{v_0^2\cdot sin^2\alpha}{g}\)
Choose the angle of attack and calculate the launch speed
or
choose the launch speed and calculate the angle of attack.
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