Question

Please respond no later than 9:00 am Saturday, June 16, 2018. Thank you.

Part 1

QUESTION How would the time of the jump and the horizontal distance traveled change if g were changed, for example if the jump could be repeated with the same initial velocity on a different planet? (Select all that apply.)  

Note: I answered b and c from one of your previous post, and got the answer wrong.

a. The time of the jump increases when g is smaller.

b. The displacement increases with increased time of the jump.

c. The time of the jump decreases when g is smaller.

d. Increasing the time of the jump has no effect on the displacement.

e. The displacement decreases with increased time of the jump.

Part 2

A bartender slides a beer mug at 1.2 m/s towards a customer at the end of a frictionless bar that is 1.3 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? 0.6181m (This is correct)

(b) What are the speed and direction of the mug at impact?

Speed____________m/s

Direction __________degree below the horizontal

Answer 1

Part 1)

a) as weknow time of flight,

Time = 2u sin(theta) /g

From above relation we can conclude that on decreasing g time offlight will increase.

b) range, R= uT

As the time increased, range also increases.

c) refer to part (a), this is a false statement.

d) refer to (b), this increasing time increases displacement.

e) referto part (b), this stement also false.

Part b)

Horizontal component of Initial speed,

ux = 1.2 m/s
Vertical component of initial speed,

uy = 0

Height of the bar H = 1.3 m

From kinematic relation (vertical motion)
H = uy*t + (1/2)gt^2
1.3 = 0 + 0.5*9.8*t^2
t^2 = 0.265
t = 0.515 s
(a)
Distance of the mug from the end of the bar = ux * t = 1.2 * 0.515 = 0.618 m
(b)
Final speed of the mug,

v = vx i + vy j
vx = ux = 1.2 m/s

vy = uy - gt
= 0 - (9.8)(0.515)
= - 5.047 m/s

Final speed v = 1.2 i - 5.047 j
Magnitude

v = Sqrt[(1.2)^2 + (-5.047)^2]

V= 5.187 m/s (Ans)

Direction,

X= tan^-1[vy/vx]
= tan^-1[5.047/1.2]
= 76.63 degrees below the horizontal   (Ans)