Question

A ball is thrown vertically upward from the ground with a velocity of

15 m/s. One second later, another ball is thrown vertically upward with a velocity of 30

m/s. how far above the ground will the two balls be at the same level.

Answer 1

$s=v_1(t+1)-g(t+1)^2/2$

$s=v_2t-gt^2/2$

$v_1(t+1)-g(t+1)^2/2=v_2t-gt^2/2\to$

$t=\frac{g/2-v_1}{v_1-v_2-g}=\frac{9.8/2-15}{15-30-9.8}=0.407\ (s)$

$s=v_2t-gt^2/2=30\cdot0.407-9.8\cdot0.407^2/2=11.4\ (m)$ . Answer