Question #8c25d

3 Answers
Gaya3
Oct 23, 2017

#+-7.544m/s#

Explanation:

#80" feet"=24.384m#
#90" feet"=27.432m#

Let's find the velocity #v# at #90# feet, on the way up.

#[#Formula : #2aS=v^2-u^2]#

#2(-9.8m/s^2)(27.432m)=v^2-(24.384m/s)^2#

#2(-9.8m/s^2)(27.432m)+(24.384m/s)^2=v^2#

#v~~color(red)(7.544m/s# which you'll get after calculating.

#therefore#On the way down, its velocity would be #color(blue)(-7.544m/s#

#-----------#

P.S. :- If you want to prove that too...

First we'll find #S_"max"# where its velocity would be #0#.

#2(9.8m/s^2)S_"max"=0^2-(24.384m/s)^2#

#S_"max"=(cancel-(24.384m/s)^2)/(cancel-2(9.8m/s^2))#

#S_"max"~~=30.34m#

Now, the distance traveled from #S_"max"# to #90# feet above the ground:
#30.336m-27.432m=2.904m#

Then we'll find the velocity #v# at #2.904m#.

#2(9.8m/s^2)(2.904m)=v^2-(0m/s)^2#

#v^2=2(-9.8m/s^2)(2.904m)#

#v=color(blue)(-7.544m/s#

A08
Oct 23, 2017

Kinematic expression is given as

#S=80t-16t^2#

Comparing it with the kinematic expression

#h=ut+1/2g t^2#, we see that
#u=80ftcdot s^-1# as given and #g=-32ftcdot s^-2#. Negative sign shows that gravity is acting against direction of initial motion.

When ball is #90ft# above the ground, #S=90ft#. Inserting given values in the applicable kinematic expression we get

#v^2-u^2=2gs#
#=>v^2-80^2=2(-32)90#
#=>v^2=80^2+2(-32)90#
#=>v^2=6400-5760#
#=>v=+-25.3ftcdot s^-1#, rounded to one decimal place

Keeping in view the #+ve# direction of #y#-axis
Velocity of the ball on its way up
#=25.3ftcdot s^-1#, rounded to one decimal place.

Velocity of the ball on its way down
#=-25.3ftcdot s^-1#, rounded to one decimal place

Michael
Oct 23, 2017

#sf(v=+25.3color(white)(x)"ft/s")# upwards.

#sf(v=-25.3color(white)(x)"ft/s")# downwards.

Explanation:

There are two instances when the ball is 90 ft above the ground.

#sf(90=80t-16t^2)#

#:.##sf(16t^2-80t+90=0)#

Applying the quadratic formula we get:

#sf(t=(80+-sqrt(6400-(4xx16xx90)))/(32))#

#sf(t=(80+-sqrt(640))/(32))#

#sf(t=(80+-25.3)/(32))#

#sf(t_1=1.71color(white)(x)s)#

#sf(t_2=3.29color(white)(x)s)#

We know that:

#sf(s=80t-16t^2)#

To get the velocity at time t we take the 1st derivative:

#sf(v=(ds)/(dt)=80-32t)#

After #sf(t_1)# seconds we get:

#sf(v=80-(32xx1.71)=+25.3color(white)(x)"ft/s")#

The + sign means the velocity is vertical.

After #sf(t_2)# seconds we get:

#sf(v=80-(32xx3.29)=-25.3color(white)(x)"ft/s")#

The - sign means the ball is falling to earth.

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