Question #101284
A toy rocket, mass 1.0 kg, blasts up at a 45o angle from ground level with a kinetic energy of 37 J. To what maximum vertical height does it rise? Round to the nearest tenth of a meter. (Hint: what portion of KE will convert into Ug?)
1
Expert's answer
2020-01-14T09:18:36-0500

The kinetic energy of a rocket

KE=mv22{\rm KE}=\frac{mv^2}{2}

So, the initial speed of a rocket


v=2KE/m=2×37/1.0=8.6m/sv=\sqrt{2{\rm KE}/m}=\sqrt{2\times 37/1.0}=8.6\:\rm m/s

The vertical component of the initial speed


vy=vsinθ=8.6×sin45=6.1m/sv_y=v\sin\theta=8.6\times \sin 45^{\circ}=6.1\:\rm m/s

The maximum vertical height


h=vy22g=6.122×9.8=1.9mh=\frac{v_y^2}{2g}=\frac{6.1^2}{2\times 9.8}=1.9\:\rm m

Answer: 1.9m1.9\:\rm m


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