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.