# Question 5ced9

Oct 22, 2017

(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,

$v = 12 m . {s}^{- 1}$

$a = g \approx 10 m . {s}^{- 1}$

$v = u + a t$

$\Rightarrow 12 = 6 + 10 t$

$t = 0.6 s$

(b)
$u = 12 m . {s}^{- 1}$

According to question,

$v = 24 m . {s}^{- 1}$

$a = g \approx 10 m . {s}^{- 1}$

$v = u + a t$

$\Rightarrow 24 = 12 + 10 t$

$t = 1.2 s$

Oct 22, 2017

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

$\left(144 - 36\right) {m}^{2} / {s}^{2} = {\left(a \cdot t\right)}^{2}$

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

$108 {m}^{2} / {s}^{2} = {\left(9.8 \frac{m}{s} ^ 2\right)}^{2} \cdot {t}^{2}$

${t}^{2} = \left(\frac{108 {m}^{2} / {s}^{2}}{96.1 {m}^{2} / {s}^{4}}\right) = 1.12 {s}^{2}$

$t = 1.06 s$

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#

$\left(576 - 36\right) {m}^{2} / {s}^{2} = {\left(a \cdot t\right)}^{2}$

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

$540 {m}^{2} / {s}^{2} = {\left(9.8 \frac{m}{s} ^ 2\right)}^{2} \cdot {t}^{2}$

${t}^{2} = \left(\frac{540 {m}^{2} / {s}^{2}}{96.1 {m}^{2} / {s}^{4}}\right) = 5.62 {s}^{2}$

$t = 2.37 s$