A toy rocket is driven vertically upwards from the origin at rest, and it rises with an acceleration of #a=(9-5t)g ms^-2# for the first 2 secs and thereafter an acceleration of #a=-g ms^-2#. How to find the max speed and max height reached?

1 Answer
A08
Jan 30, 2018

Given acceleration

#a=(9−5t)#

We know that acceleration #a-=dotv#, where #v# is the velocity. Rewriting given expression

#dotv=(9−5t)#

Integrating both sides with respect to #t# we get

#int\ dotv\ dt=int\ (9−5t)\ dt#
#=>v=(9t−5t^2/2+C)#
where #C# is a constant of integration.

It is given that initial velocity at #t=0# is #0#, #:.C=0#. Expression for #v# becomes

#=>v=(9t−5/2t^2)# .......(1)

Velocity will be maximum when acceleration is #=0#.

#=>0=(9−5t)#
#=>t=9/5=1.8\ s#

From (1) #v_max=9xx1.8-5/2xx(1.8)^2#

#v_max=16.2-8.1=8.1\ ms^-1#

Velocity at the end of #2 \ s#, from (1)

#v(2)=(9xx2−5/2xx2^2)#
#=>v(2)=18−10=8\ ms^-1#

We know that #v-=doth#, where #h# is the height of rocket. Rewriting expression (1) in terms of #h# and integrating both sides with respect to #t# we get

#int\ doth\ dt=int\ (9t−5/2t^2)\ dt#
#h=(9t^2/3−5/2t^3/3+C_1)#
where #C_1# is a constant of integration.

It is given that initial height at #t=0# is #0#, #:.C_1=0#. Expression for #h# becomes

#h=3t^2−5/6t^3#

Height reached at #t=2\ s#

#h(2)=3(2)^2−5/6(2)^3#
#h(2)=12−20/3#
#h(2)=5.bar3\ m#

At this point of time acceleration changes to a constant value #=-g\ ms^-2#. Peak height above #h(2)# will be reached when velocity #=0#. Now using the kinematic expression

#v^2-u^2=2gh#

Inserting given and calculated values we get

#(0)^2-(8)^2=2(-g)h_"peak"#
#=>h_"peak"=32/g\ m#

Now #h_max=h(2)+h_"peak"#

#h_max=(5.bar3+32/g)\ m#

Taking value of #g=9.8\ ms^-2#

#h_max=(5.bar3+32/9.8)#
#h_max=8.60\ m#, rounded to two decimal places.

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