Chapter 4, Problem 54A

a.

To determine

Diver velocity at the instant of hiting the water.

Answer to Problem 54A

When diver hits the water his velocity will be 14 m/sec.

Explanation of Solution

Given :

Mass of diver $m=65\text{ kg}$

Height of tower $h=10\text{ m}$

Formula used :

According to third equation of motion,

  $v^2=u^2+2gh$

Where,

u is initial velocity

v is final velocity

g is acceleration due to gravity

h is height.

Calculation:

When the diver jumps off his initial velocity is $u=0\text{ m/sec}$

Acceleration due to gravity is $g=9.8{\text{ m/sec}}^2$

Now, substitute the value of u, g and hin equation of motion.

  $\begin{array}{l}{v}^2=0^2+2\times 9.8\times 10\\ v^2=196\\ v=14\text{ m/sec}\end{array}$

Conclusion:

Therefore, when diver hits the water his velocity will be 14 m/sec.

b.

To determine

Net force exerted by the water.

Answer to Problem 54A

Net force exerted by water is 3185 N.

Explanation of Solution

Given :

Mass of diver $m=65\text{ kg}$

Distance travelled $s=2\text{ m}$

Formula used :

According to third equation of motion,

  $v^2=u^2+2gh$

Where,

u is initial velocity

v is final velocity

g is acceleration due to gravity

h is height.

Also, the force is

  $F=ma$

Where,

m is mass

a is acceleration

Calculation:

When the diver jumps off his initial velocity $u=14\text{ m/sec}$

After travelling a distance of 2 m diver finally stops

Therefore, final velocity of diver is $v=0\text{ m/sec}$

Substitute the value of u, v and s.

  $\begin{array}{c}{0}^2={14}^2+2\times a\times 2\\ 4a=-{14}^2\\ a=\frac{-{14}^2}{4}\\ =-49{\text{ m/sec}}^2\end{array}$

Now, substitute the value of m and ain the equation of force.

  $\begin{array}{c}F=65\times -49\\ =-3185\text{ N}\end{array}$

Here, negative sign indicates that direction of force is in opposite direction of motion.

Conclusion:

Therefore, the force exerted by water in opposite direction with a magnitude of 3185 N.