Suppose you throw a stone straight up with an initial speed of 15 m s−1

Question

.

(a) (5 points) If you throw a second stone straight up 1.0 s after the first, with what speed must you throw
the second stone if it is to hit the first at a height of 11.0 m? There are two answers. Do both solutions
make physical sense?
(b) (2 points) Plot x − t and v − t graphs for both balls.
(c) (3 points) If you throw the second stone 1.30 s after the first, with what speed must you throw the second
stone if it is to hit the first at a height of 11.0 m?

Answer 1

(a) when is the first stone at 11m?
15t-4.9t^2 = 11
t = 1.21, 1.84
So, you need the 2nd stone to be at 11m one second faster.
v(t-1)-4.9(t-1)^2 = 11
v = 53.41
Repeat with t=1.84 to meet on the way down.
(b) I assume you can graph parabolas
(c) change t-1 to t-1.3