<p>i know this is a physics question, which i posted before, but does anyone know how to do it in math form? specifically for calc, by any chance? </p>
<p>Melinda took a little nosedive from her perch on top of the building, 25 feet above ground. Given that she fell as a result of a gentle tap to her noggin, how fast is Melinda traveling when she hits the ground? (acceleration due to gravity is -32 ft/sec^2). Make sure your answer is in feet/sec.</p>
<p>X = Xo + VoT + (1/2)A(T^2)
Where:
X is the final position (25ft, she fell 25ft)
Xo is the starting position (0ft)
Vo is the initial velocity (0ft/s)
T is the time it took traveling
A is acceleration (32ft/s/s)</p>
<p>You can rewrite to X = .5AT^2</p>
<p>Solve for T. That is the time it took to reach the ground. During that time interval, the object was accelerating at a constant rate of 32 ft/s/s, so after 1 second the object had a velocity of 32ft/s, after 2 seconds its velocity was 64ft/s, etc. So multiply T by 32 ft/s/s to find out what the velocity is at the instant Melinda hits the ground.</p>
<p>hmm…ok thanks…i’ll see if there’s a calculator mistake b/c the answer is supposed to be
-40…</p>
<p>no matter what i do, i can’t get the right answer of -40…</p>
<p>Yes, I realize the answer should have been -40/40 (dependent on which direction you indicate as positive) because I answered your last thread. Anyways here is the corrected version. Made a careless and illogical mistake.</p>
<p>d’(t) = v(t)
v’(t) = a(t)
integral of a(t) = v(t)
integral of v(t) = d(t)</p>
<p>a(t) = -32
v(t) = -32t
d(t) = -16t^2</p>
<p>-25 = -16t^2
25/16 = t^2
t = 5/4
t = 1.25</p>
<p>v(t) = -32t
v(t) = -32(1.25)
v(t) = -40 ft/s</p>
<p>thanks so much…haha i actually just figured that out a little before after looking up the right equation too. lol</p>
<p>sry but can u help me with one last question please? This question is from a little math story question so the words might not make sense…</p>
<p>“While you were walking the line for the cop, I was hiding in the bushes with my radar gun. I made a graph of your velocity in feet per minute during your little 3-minute jaunt. At time=0, you were backed up against the bumper of your car. Answer the question to 3 decimal places.”</p>
<p>Question: At what time was your instantaneous velocity the same as your average velocity over the first minute? </p>
<p>Thanks again!!</p>