jonathan and jennifer are testing a physics experiment. jonathan drops a penny from the rooftop of a 160 foot tall building into a pool below on ground level. jennifer, who is next to him, throws her penny straight down at 48 feet per second at the same instant. by how many seconds does her penny beat jonathan's into the water?

Question

mathstudent55 · Answer

$d = v_0 t + \dfrac{1}{2}gt^2$

mathstudent55 · Answer

Use the given info each time, and solve for t each time.

mathstudent55 · Answer

For Jonathan, v0 = 0
For Jennifer, v0 = 48 ft/s

anonymous · Answer

what;s the gravity?

mathstudent55 · Answer

$d = v_0 t + \dfrac{1}{2}gt^2$

Jonathan: 

$-160 = \dfrac{1}{2}(-32.2)t^2$

$t = 3.15~s$

Jennifer: 

$-160 = -48t + \dfrac{1}{2}(-32.2)t^2$

$16.1t^2 + 48t - 160 = 0$

$t = \dfrac{-48 \pm \sqrt{48^2 - 4(16.1)(-160)}}{2(16.1)}$

$t = 1.996$   or  $t = -4.978$

Discard the negative solution, so t = 1.996 for Jennifer.

mathstudent55 · Answer

Now find the difference between the times.