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Question Number 180917 by anne1344 last updated on 19/Nov/22

Answered by mr W last updated on 19/Nov/22

$$c)$$$$h_{max}=\frac{{\left(v\sin \theta \right)}^2}{2g}=\frac{{\left(15\times \sin 20°\right)}^2}{2\times 9.81}=1.34m$$$$d)$$$$t=\frac{v\sin \theta }{g}=\frac{15\times \sin 20°}{9.81}=0.52s$$

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