Answer to Question #90886 in Mechanics | Relativity for Carmen

Question

A stone is thrown vertically upward at a speed of 29.10 m/s at time t=0. A second stone is thrown upward with the same speed 2.190 seconds later. At what time are the two stones at the same height?

Answer 1

$$h=vt-0.5gt^2$$

$$h=v(t-2.19)-0.5g(t-2.19)^2$$

Thus,

$$vt-0.5gt^2=v(t-2.19)-0.5g(t-2.19)^2$$

$$v(-2.19)-0.5g(-2(2.19)t+2.19^2)=0$$

$$29.1(-2.19)-0.5(9.81)(-2(2.19)t+2.19^2)=0$$

$$t=4.061\ s$$