Veronica angrily throws her engagement ring? (See below)

Veronica angrily throws her engagement ring straight up from the roof of a building, 12.0 m above the ground, with an initial speed of 6.00 m/s. Air resistance may be ignored. For the
motion from her hand to the ground, what are the magnitude and
direction of
a) the average velocity of the ring?
b) the average acceleration of the ring?
c) Sketch a-t, v-t, and y-t graphs for the motion of the ring.

1 Answer
Jul 5, 2018

Let us us consider origin #O# at the ground level just below the point where Veronica threw her ring.

Roof of the building where the girl stands as #R#
Highest point where velocity of ring becomes #=0# as #T#.
Upward direction is taken as positive. Direction of acceleration due to gravity is downwards, therefore #g=-10\ ms^-2#

Total time #t# taken by the ring to reach the ground level. (from #R# to #O)#.
Applicable kinematic expression is

#h = h_0+ut + 1/2 g t^2# .......(1)

Inserting given values we get

#0 = 12.0+6.00t + 1/2 (-10) t^2#
# => 5 t^2-6.00t -12.0=0#

This quadratic can be solved using standard method. I used in built graphics calculator.
my comp

We get solutions as #t=-1.061and 2.261#
Ignoring the negative root as time can not be negative we get

#t=2.261\ s#

(a) #"Average velocity" = "Total displacement"/ "total time"#
#"Average velocity" = (-12)/ 2.261#
#=>"Average velocity"= -5.31\ ms^-1#

(b) #"Average acceleration" = "Change in velocity"/ "time interval for change"#
#"Average acceleration"= ("final velocity" - "initial velocity")/"total time"# .....(2)

Final velocity, the velocity of ring as it hits the ground can be calculated with the kinematic expression

#v = u + at# (from #R# to #O#)
#v = 6.00 + (-10)(2.261)#
#v= -16.61\ ms^-1#

#:.# from (2)
#"Average acceleration"= (-16.61 - 6.00)/2.261=-10\ ms^-2#

(c) For sketching graphs we need time taken by the ring to come to halt momentarily at #T# using the kinematic expression

#v = u + at#
#0 = 6.00 + (-10)t_T#
# =>t_T=0.6\ s#

Now height of point #T# is calculated using (1)

#h_T = 12.0+6.00xx0.6 + 1/2 (-10) (0.6)^2#
#h_T = 10.56\ m#

With these equations and values draw various sketches.