A ball is thrown upward from the ground with an initial speed of 49 m/s; at the same instant, another ball is dropped from a building 16 m high. After how long will the balls be at the same height?
Use the displacement equations
Let the first ball have velocity u(t), displacement x(t), given:
x(0)=0
u(0)=49m/s.
x(t) = u(0)t - (1/2) gt^2
Let the second ball have velocity v(t), displacement y(t), given:
y(0)=16m
v(0)=0.
y(t) = y(0) - (1/2) gt^2
Find t such that x(t)=y(t)
.: u(0) t = y(0)