A body is dropped from a height of 100 yards. After falling for 2 seconds a second body is projected vertically upward from a point on the earth directly below and with a velocity of 40 yd/sec (yards per second). Find the time and height at which the two bodies meet.
My logic is as follows:
$s_1=-16t^2+100$ for the object being dropped.
The next step I think is more complicated, I assume that:
when $t=2, v_2=-32t+c, 40=-64+c, c=104$
Therefore, $s_2=-16t^2+104t$
When $s_1=s_2$,
$-16t^2+100=-16t^2+104t, t=\frac{100}{104}$
This is the wrong answer however, could somebody please explain how to do this correctly?
You project the object after $2$ seconds so it travels for $t-2$ seconds.
So the distance traveled by object projected = $40(t-2) - \frac{g}{2}(t-2)^2$
Distance traveled by dropped object $= \frac{g}{2}t^2$ where $g$ is the gravity.
So total distance $100 = \frac{g}{2}t^2 + 40t - 80 - \frac{g}{2}t^2+2gt-2g$
$(2g+40)t = 180+2g \implies t = \frac{90+g}{20+g}$.