<p>hey can someone help me with this problem please? thanks in advance! </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>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>You should probably be familiar with the equation for position: y= 1/2<em>g</em>t^2 + vo*t + yo</p>
<p>y= height at time t
g = gravity
vo = initial velocity
yo = initial height</p>
<p>y’ = derivative of position = velocity = v</p>
<p>y’ = v = gt + vo</p>
<p>so… use position equation to find the time it takes to hit the ground. plug that time into velocity equation to find velocity at time t. </p>
<p>The “gentle tap” part of the question is another way of saying initial velocity is zero.</p>
<p>FYI you can get the position and velocity equations starting with the differential equation:</p>
<p>dv/dt = g</p>
<p>and initial conditions</p>
<p>v(t=0) = vo
y(t=0) = yo</p>