Question

At time t=0, rock A is dropped from rest from a height of 90 m. At the same instant, rock B is launched straight up from the ground level with an initial speed of 30 m/s. Write an equation of motion for rock A and B, giving its position at all time.

Answer 1

Answer:

rock A: $y=90-1/2*g*t^2$

rock B: $y=30*t-1/2*g*t^2$

g=9.81m/s^2

Explanation:

Kinematics equation:

$y=y_{o}+v_{oy}*t+1/2*a*t^2$

in our case the acceleration is the gravity and it has a negative direction.

a=-g

rock A, yo=90m, Voy=0m/s:

$y=90-1/2*g*t^2$

rock B, yo=0m, Voy=30m/s:

$y=30*t-1/2*g*t^2$