Answer to Question #234123 in Physics for Cat

Question

a ball is thrown upward from the ground at an angle of 60° with the horizontal with a velocity of 6m/s. a)find its vertical postion with respect to the ground when it has travelled a distance of 2m horizontally. b) find its velocity(upward or downward) at this point

Answer 1

Given:

$\theta=60^{\circ}$

$v_0=6\:\rm m/s$

$x=2\:\rm m$

(a) the equation of trajectory of a ball is given by

$$y=x\tan\theta-\frac{gx^2}{2v_0^2\cos^2\theta}$$

So

$$y=2\tan 60^{\circ}-\frac{9.8*2^2}{2*6^2*\cos^260^{\circ}}=1.3\:\rm m$$

(b)

$$x=v_0\cos\theta*t$$

$$t=\frac{x}{v_0\cos\theta}=\frac{2}{6*\cos60^{\circ}}=0.67\:\rm s$$

The components of velocity vector are

$$v_x=v_0\cos\theta=6*\cos60^{\circ}=3\:\rm m/s$$

$$v_y=v_0\sin\theta-gt=6*\sin60^{\circ}-9.8*0.67\\ =-1.3\:\rm m/s$$

The velocity is directed downward.