Question

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?

Answer 1

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