Question

A ball is thrown at 25 m/s, 20 degrees above the horizontal. What is the magnitude of the resultant velocity at a time of t = 0.35 s?

Answer 1

$V_y=V_0sin\theta= 25\sin 20= 22.8$

$V_x=V_0\cos\theta= 25\cos 20\degree= 10.2$

$V_y=V_0+a_yt= 22.8-9.8(0.35)=19.37$

$\tan \theta=\frac {19.4}{10.2}$

$\theta=62.2\degree$

$magnitude= 19.4m/s$ $62.2\degree$ below the horizontal $(62.2 \degree from X-axis.)$