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Question Number 155035 by peter frank last updated on 24/Sep/21

Answered by mr W last updated on 24/Sep/21

$$h_{max}=\frac{u_0^2{\sin }^2\theta }{2g}$$$$p.e.ath=\frac{h_{max}}{2}=\frac{u_0^2{\sin }^2\theta }{4g}is$$$$m_{0}gh=\frac{m_0{u}_0^2{\sin }^2\theta }{4}$$

Commented by peter frank last updated on 25/Sep/21

$$\mathrm{thank}\mathrm{you}$$

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