Answer to Question #269887 in Quantum Mechanics for Vloky

Question

An object is launched at a velocity of 20m/s in a direction making an angle of 25° in a 50 meter elevation. How long is the object in the air? What is the range of the object?

Answer 1

$1. \ v_y=v_0\sin25°-gt_1\to t_1=v_0\sin25°/g=20\cdot\sin25°/9.8=0.86\ (s)$

$H=50+v_0^2\sin^225°/(2g)=50+20^2\cdot\sin^225°/(2\cdot9.8)=53.65\ (m)$

$H=gt_2^2/2\to t_2=\sqrt{2H/g}=\sqrt{2\cdot 53.65/9.8}=3.3\ (s)$

$t=t_1+t_2=0.86+3.3\approx4.2\ (s)$ . Answer

$2.\ x=v_xt=v_0\cos25°\cdot t=20\cdot\cos25°\cdot4.2\approx76.1\ (m)$ . Answer