Answer to Question #177047 in Physics for James Masela

Question #177047

The ball is thrown at an angle to the horizon, fell to the ground after 4 seconds. The maximum speed of the ball is twice it’s minimum speed. Find the maximum height of the ball


1
Expert's answer
2021-03-31T09:32:00-0400
vxv0=12=cosαα=60°\frac{v_x}{v_0}=\frac{1}{2}=\cos \alpha\\\alpha=60\degree

T=2v0sin60g4=2v0sin609.8v0=22.6msT=\frac{2v_0\sin{60}}{g}\\4=\frac{2v_0\sin{60}}{9.8}\\v_0=22.6\frac{m}{s}

H=v02sin2602gH=22.62sin2602(9.8)=19.5 mH=\frac{v_0^2\sin^2{60}}{2g}\\ H=\frac{22.6^2\sin^2{60}}{2(9.8)}=19.5\ m


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