Question

an object X was drop from a height 15.28m above the ground. just as that time, an object Y was short vertically up from the ground with a velocity of 18m/s.object X fell through a distance of 3.6m and meet object Y.
a. calculate the velocity of the two bodies when they meet
b. Calculate when the two bodies meet

Answer 1

b)

$$h=0.5gt^2$$

$$3.6=0.5(9.8)t^2$$

$$t=0.857\ s$$

a)

$$v_x=-(9.8)(0.857)=-8.4\frac{m}{s}$$

$$H-h=v_{y0}t-0.5gt^2$$

$$15.28-3.6=(0.857)v_{y0}-0.5(9.8)(0.857)^2$$

$$v_{y0}=17.83\frac{m}{s}$$

$$v_y=17.83-(9.8)(0.857)=9.4\frac{m}{s}$$