Question

Question #5ced9

Answer 1

(a) 0.6 s

(b) 1.2 s

Explanation:

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`

Answer 2

a) `t = 1.06 s`
b) `t = 2.37 s`

Explanation:

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`