Answer to Question #176696 in Physics for Francine

Question #176696

A projectile has an initial launch speed of 18.0 m/s. If it is desired that it strikes a

target 31.0 m away at the same elevation, what should be the projection angle?


1
Expert's answer
2021-03-29T16:47:29-0400

We can find the projection angle from the definition of the range:


R=v02sin2θg,R=\dfrac{v_0^2sin2\theta}{g},sin2θ=Rgv02,sin2\theta=\dfrac{Rg}{v_0^2},2θ=sin1(Rgv02),2\theta=sin^{-1}(\dfrac{Rg}{v_0^2}),2θ=sin1(31 m9.8 ms2(18 ms)2)=70,2\theta=sin^{-1}(\dfrac{31\ m\cdot9.8\ \dfrac{m}{s^2}}{(18\ \dfrac{m}{s})^2})=70^{\circ},θ=702=35.\theta=\dfrac{70^{\circ}}{2}=35^{\circ}.

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