The equation of trajectory of the ball when it moves as projectile
y=y0+tan(θ)⋅x−2v02cos2(θ)g⋅x2In our case we have an equation
3.05=1.95+tan(39∘)⋅6.75−2v02cos2(39∘)9.8⋅6.752So, the initial velocity
v0=9.2m/sThe maximum height
ymax=y0+2gv02sin2(θ)
=1.95+2×9.89.22sin2(39∘)=3.66mSo, the ball’s maximum potential energy during the throw
Emax=mgymax
=0.625kg×9.8N/kg×3.66m=22.4J
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