Question

A 70-kg diver jumps off a 10.2-m tower.

(a) Find the diver's velocity when he hits the water.
?m/s

(b) The diver comes to a stop 2.0 m below the surface. Find the net force exerted by the water.
?N

Answer 1

Let's start with (a). We can look at solving using conservation of energy, or kinematics. Do you have an approach you want to use?

Answer 2

Kinematics

Answer 3

Ok. So we know vertically this will be a free-fall problem. Which gives us the known acceleration due to gravity a = 9.8 m/s^2. I will use the coordinate system where up is positive and down is negative so the acceleration would be - 9.8 m/s^2.

The problem statement doesn't mention jumping upward so we will assume their initial vertical speed is 0 m/s. We also know they are coming off the 10.2m tower, so their vertical displacement will be 10.2m downward, or -10.2m

viy = 0 m/s
ay = -9.8 m/s^2
\[\Delta y = -10.2m\]
vfy = ?

So we need a kinematics equation with initial velocity, final velocity, acceleration, and displacement.

\[v_{fy}^{2}=v_{iy}^{2} + 2a \Delta y\]

Answer 4

How's it going?

Answer 5

I'm confused by the 2aΔy part of the equation

Answer 6

Final Velocity = 0 + 2(-9.8) (-10.2) Would this be right?

Answer 7

yes, but remember that is the final velocity squared. so to solve you should take the square root of both sides.

Answer 8

because you are taking the square root, you then should go back and check which answer would make sense a + final velocity or a - final velocity. The - makes sense with our coordinate system.

Answer 9

Okay thank you! But how would you do part B?

Answer 10

For part (b) you can use that final velocity as it hits the water, as your initial velocity for part be, with the diver then coming to rest (velocity final = 0 m/s) after moving 2 meters downward (-2m).

You would use the same kinematics equation but this time solve for the acceleration a.

Once you have the acceleration you would use Newton's 2nd law to find Force:
\[\sum F = ma\]

Answer 11

Which will end you with \[\Sigma F = 3498.943 N\] Thank you so much!