<p>hey guys, can u help me with one physics question please?</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>Thankss:)</p>
<p>d = -25 ft -> negative because she’s going down
a = -32 ft/s^2
v(i) = 0
v(f) = ?</p>
<p>v(f)^2 = v(i)^2 + 2 * -32 * -25 -> simplifies to v(f) = 40 ft/s or -40 ft/s.</p>
<p>Since she is going down, I think v(f) should be -40 ft/s. I haven’t taken physics in over a year, but this makes sense.</p>
<p>Yi = 25 feet
Yf = 0 feet
Vi = 0 m/s
Vf = ?
a = -32 ft/sec^2</p>
<p>(Vf)^2 = (Vo)^2 + 2a/\y
Vf^2 = 0 + 2(-32)(0 - 25)
Vf^2 = 1600
Vf = 40 or -40</p>
<p>Since up is indicated as positive Vf = -40 ft/s.</p>
<p>^The distance is 25 ft so vf^2 = 1600 –> vf = 40/-40. You are correct (I think) in that the final velocity should be negative.</p>
<p>^Ya, I corrected it I plugged in 20 instead of 25 by accident. Thanks though.</p>
<p>^The acceleration due to gravity in this problem is -32 ft/s^2 whether the girl is traveling upwards or downwards. </p>
<p>If you indicate upwards as positive then the answer will be -40 ft/s.<br>
If you indicate downwards as positive then the answer will be 40 ft/s.</p>
<p>You can also solve the problem through conservation of energy:</p>
<p>Gravitational Potential Energy at the beginning = Kinetic Energy at the end.</p>
<p>mgh = ½mv²
v = √2gh
v = √2(-32 ft/sec²)(-25 ft)
v = 40 ft/sec or -40 ft/sec, depending on whether the positive direction is up or down.</p>
<p>o ok…thanks so much! oop i didin’t even realize this was the SAT not SAT subject tests forum. lols</p>
<p>sry but can u guys help me with one last question please? This question is from a little math story activity sheet 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>