Question #5ced9

As the fish falls with constant acceleration i.e #g#, we can apply the equations of motion for both the questions.

(a)

#u=6 m.s^(-1)#

According to question,

#v=12 m.s^(-1)#

#a=g~~10 m.s^(-1)#

#v=u+at#

#rArr 12=6+10t#

#t=0.6 s#

(b)
#u=12 m.s^(-1)#

According to question,

#v=24 m.s^(-1)#

#a=g~~10 m.s^(-1)#

#v=u+at#

#rArr 24=12+10t#

#t=1.2 s#

Speed is the magnitude of velocity. The fish will continue to have its horizontal speed at 6.0 m/s. Speed will be the sum of its increasing vertical speed and its constant horizontal speed.

Its increasing vertical velocity will be a*t downward. So its velocity, will be the vector sum of its vertical and horizontal velocities. Its speed can be found using Pythagoras
#(speed)^2 = sqrt((6.0 m/s)^2 + (a*t)^2#

a) We want the time at which the speed is 12 m/s, so

#(12 m/s)^2 = sqrt((6.0 m/s)^2 + (a*t)^2#

#(144 - 36) m^2/s^2 = (a*t)^2#

Using 9.8 m/s^2 as the value of a,

#108 m^2/s^2 = (9.8 m/s^2)^2*t^2#

#t^2 = ((108 m^2/s^2)/(96.1 m^2/s^4)) = 1.12 s^2 #

#t = 1.06 s#

b) We now want the time at which the speed is 24 m/s, so

#(576 m/s)^2 = sqrt((6.0 m/s)^2 + (a*t)^2#

#(576 - 36) m^2/s^2 = (a*t)^2#

Using 9.8 m/s^2 as the value of a,

#540 m^2/s^2 = (9.8 m/s^2)^2*t^2#

#t^2 = ((540 m^2/s^2)/(96.1 m^2/s^4)) = 5.62 s^2 #

#t = 2.37 s#