The 10 kg dancer leaps into the air with an initial velocity of 5 m/s at angle of 45° from the floor. how far will she travel in the air horizontally before she lands on the ground again?

Respuesta :

Donek
[tex]Given:\\m=10kg\\v_0=5 \frac{m}{s} \\ \alpha =45^\circ\\g=10 \frac{m}{s^2} \\\\Find:\\r=?\\\\Solution\\\\r= \frac{v_0^2\sin 2 \alpha }{g} \\\\r= \frac{(5 \frac{m}{s})^2\sin2\cdot45^\circ}{10 \frac{m}{s^2} } = \frac{25 \frac{m^2}{s^2} }{10 \frac{m}{s^2} } =2,5m\\\\\\proof\;that \;r=\frac{v_0^2\sin 2 \alpha }{g} \;in\;appendix[/tex]
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