Question

A bartender slides a beer mug at 1.7 m/s towards a customer at the end of a frictionless bar that is 1.0 m tall. The customer makes a grab for the mug and misses, and the mug sails off the end of the bar. (a) How far away from the end of the bar does the mug hit the floor? m (b) What are the speed and direction of the mug at impact?

Answer 1

Answer:

Explanation:

Given

velocity of mug with which it leaves the bar is 1.7 m/s

Also

height of bar=1 m

Considering motion in vertical direction

$s=u_yt+\frac{gt^2}{2}$

here $u_y=0$

$1=0+\frac{9.81\times t^2}{2}$

$t=\sqrt{\frac{2}{9.81}}$

t=0.451 s

so horizontal distance traveled is

$R_x=ut+\frac{at^2}{2}$

here a=0

$R_x=1.7\times 0.451=0.766 m/s$

speed of mug will be combination of horizontal and vertical velocity

$v_y=u+gt$

$v_y=0+9.81\times 0.451=4.42 m/s$

$v_x=1.7 m/s$

thus $v_{net}=\sqrt{v_x^2+v_y^2}$

$v_{net}=\sqrt{4.42^2+1.7^2}$

$v_{net}=\sqrt{22.46}=4.74 m/s$

for direction

$tan\theta =\frac{4.42}{1.7}=2.6$

$\theta =69$

$\theta$is with x axis in clockwise sense