h1 = Vo*t + 0.5g*t^2
h1 = 16*0.65 - 4.9*0.65^2 = 8.33 m. Head
start.
V1 = Vo + g*t = 16 - 9.8*0.65=9.63 m/s.
h2 = Vo*t + 0.5g*t^2=8.33+V1*t+0.5g*t^2
27*t - 4.9t^2 = 8.33 + 9.63t - 4.9t^2
27t - 9.63t - 4.9t^2 + 4.9t^2 = 8.33
17.37t = 8.33
t = 0.48 s.
h = 27*0.48 - 4.9*0.48^2 = 11.83 m Above
launching point.
ll is thrown vertically upward with an
initial speed of 16 m/s. Then, 0.65 s later, a
stone is thrown straight up (from the same
initial height as the ball) with an initial speed
of 27 m/s.
How far above the release point will the ball
and stone pass each other? The acceleration
of gravity is 9.8 m/s
2
.
Answer in units of m
1 answer
2 answers