# Diving Board Jump: Analyzing Velocity and Height

• Jabababa
In summary: This means the diver is moving downwards at 7.87 m/s as she enters the water.In summary, a diver jumps off a 3.00 meter high diving board with an initial velocity of 1.75 m/s vertically upward. Using the equation v^2=v0^2+2ad, with a negative value for acceleration and displacement, the diver's velocity when she reaches the water is 7.87 m/s in the downward direction.

## Homework Statement

A diver jumps off a 3.00 meter high diving board with an initial velocity of 1.75 m/s vertically upward.

a) What is the diver's veolcity when she reaches the water? (assume the surface of the water is 3.00 meters below the board.)

b) What is the highest point the diver reaches above the water?

## The Attempt at a Solution

a) Since we know the height, initial velocity and acceleration, we use V^2= Vo^2 + 2ad
v=square root of vo^2 + 2ad
v = square root of 1.75^2 + 2(9.80)(3)
v= 7.86527 m/s = 7.87m/s

b)we use the same equation but this time we only calculate the diver going up, the maximum height of the the jump.
V^2 = Vo^2 + 2ad
d= V^2-Vo^2/2a
d= Vo^2/2a
d= 0.3125m

The maximum height = diver's jump + the 3.00m below the diving board.
so the answer is 3.3125m is the highest point the diver reaches above the water.

Can anyone check see if what i did is right? Thank you.

A useful check for this is that the velocity she hits the water should equal what she would have if she fell in freefall starting with zero velocity from the max height attained.

Your answer to (a) looks correct.

thank you NascentOxygen

Is b) correct as well? The logic behind it seems correct to me. Finding the max height the diver jumps plus the 3.00m below the diving board seems reasonable to me.

I am not sure what you have done in part b) .

For maximum height above diving board use the relation v2=u2+2as where v=0,u=1.75m/s,a=-9.8 m/s2.

Tanya Sharma said:
I am not sure what you have done in part b) .

For maximum height above diving board use the relation v2=u2+2as where v=0,u=1.75m/s,a=-9.8 m/s2.

Yep, thank you for clarifying. Thats exactly what i did, and the quest ask what is the max height above the water, so the max height above the board added to the 3.00m from the board to the water.

Maybe you confused at what i did (or you didnt read), i just used different variables than you, but the same equation. Anyways, thanks a lot!

You have made error while calculating 'd' .The correct answer should be 3.15625 m.

omg thank you! I feel really stupid now :(

but it was 2am when i did it, anyways thanks!

Jabababa said:

## Homework Statement

A diver jumps off a 3.00 meter high diving board with an initial velocity of 1.75 m/s vertically upward.

a) What is the diver's veolcity when she reaches the water? (assume the surface of the water is 3.00 meters below the board.)

a) Since we know the height, initial velocity and acceleration, we use V^2= Vo^2 + 2ad
v=square root of vo^2 + 2ad
v = square root of 1.75^2 + 2(9.80)(3)
v= 7.86527 m/s = 7.87m/s

This response may seem irritatingly pedantic but hopefully it is useful.

You have taken the correct steps mathematically - and in fact your result is almost correct - but you need to be careful. The first thing I will say is that you have not found the diver's velocity but rather her speed. As you have it the "velocity" is a positive number which would suggest that the diver hits the water while traveling upwards (in the direction of her initial jump). This might be irritating but any teacher will take marks off for this kind of oversight. The distinction between speed and velocity is one of the most important ideas most physics teachers try to convey at this level.

The second thing I will write out explicitly for you. You've already solved the first part so I'll show you my solution and highlight the primary difference. You might very well be aware of this difference, but it's very important that you are aware of it. As you know, this problem can be solved by the following equation:

##v_{f}^{2} = v_{i}^{2} + 2a Δd##

I will now become highly pedantic in my approach to solving this problem. I recommend a similar approach to solving physics problems, especially when you're first learning how to approach them. This method is termed the GRASP method.

Given: (write out what we know)
$$v_{i} = 1.75 \frac{m}{s}$$ (in the upward direction)
$$a = - 9.8 \frac{m}{s^{2}}$$ (in the downward direction)
$$Δ d = -3 m$$

Required: (write out what we want to find)
##v_{f}##

Assess: (write out useful equations
##v_{f}^{2} = v_{i}^{2} + 2a Δd##

Solve:
##v_{f}^{2} = v_{i}^{2} + 2a Δd##
##v_{f} = (v_{i}^{2} + 2a Δd)^{\frac{1}{2}}##
##v_{f} = ((1.75)^{2} + 2(-9.8) (-3))^{\frac{1}{2}}## This is the difference - both my a and d are negative - so the negatives cancel each other out - but it's important to aware that they are as such
##v_{f} = 7.87 \frac{m}{s}##

I know that probably seems a silly reason for me to go through the whole problem again, but it's very important to be aware of whether or not each variable you consider is positive or negative because we're dealing with vectors.

Paraphrase:
The velocity of the diver as she enters the water is 7.87 m/s in the negative direction.

Last edited: