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Write a quadratic equation that has the projectile reach a maximum height of 45 meters

 May 2, 2022
 #1
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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.

laugh  !

 May 2, 2022

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