Question

a child drops a toy at rest from a point 4m above the ground at the same instant her friend throws a ball upward at 6m/s frrom a point 1m above the ground. at what distance above the ground do the ball and the toy cross paths?

Answer 1

d1 + d2 = 4-1 = 3 m.
(0.5g*t^2) + (Vo*t+0.5g*t^2) = 3 m.
4.9t^2 + 6t+(-4.9t^2) = 3
6t = 3
t = 0.5 s.

d1 = ho - 0.5g*t^2
d1 = 4 - 4.9*0.5^2 = 4 - 1.225=2.775 m.
Above gnd.

d2 = 1 + (6*0.5 + (-4.9)*0.5^2
d2 = 1 + (3 + (-1.225) = 1 + 1.775=2.775
m. Above gnd.

Therefore, the ball and toy cross paths
at 2.775 m. above gnd.