Answer to Question #234123 in Physics for Cat

Question #234123

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


1
Expert's answer
2021-09-07T09:50:24-0400

Given:

θ=60\theta=60^{\circ}

v0=6m/sv_0=6\:\rm m/s

x=2mx=2\:\rm m


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

y=xtanθgx22v02cos2θy=x\tan\theta-\frac{gx^2}{2v_0^2\cos^2\theta}

So

y=2tan609.822262cos260=1.3my=2\tan 60^{\circ}-\frac{9.8*2^2}{2*6^2*\cos^260^{\circ}}=1.3\:\rm m

(b)

x=v0cosθtx=v_0\cos\theta*t

t=xv0cosθ=26cos60=0.67st=\frac{x}{v_0\cos\theta}=\frac{2}{6*\cos60^{\circ}}=0.67\:\rm s

The components of velocity vector are

vx=v0cosθ=6cos60=3m/sv_x=v_0\cos\theta=6*\cos60^{\circ}=3\:\rm m/s

vy=v0sinθgt=6sin609.80.67=1.3m/sv_y=v_0\sin\theta-gt=6*\sin60^{\circ}-9.8*0.67\\ =-1.3\:\rm m/s

The velocity is directed downward.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment