Answer to Question #101284 in Physics for bob

Question

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?)

Answer 1

The kinetic energy of a rocket

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

So, the initial speed of a rocket

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

The vertical component of the initial speed

$$v_y=v\sin\theta=8.6\times \sin 45^{\circ}=6.1\:\rm m/s$$

The maximum vertical height

$$h=\frac{v_y^2}{2g}=\frac{6.1^2}{2\times 9.8}=1.9\:\rm m$$

Answer: $1.9\:\rm m$