the mass of the ball does not matter.
The height as a function of time is
h(t) = 30 + 20t - 4.9t^2
so, just find the vertex of the parabola, and solve for t when h=0. Just good old Algebra I.
A ball with a mass 0.15kg is thrown upward with an initial velocity 20m/s from the roof of a building 30m high. Neglect air resistance.
Find the max height above the ground that the ball reaches.
Assuming the ball misses the building find the time that it hits the ground
3 answers
so this is for a diff eq class. how did you use differential equations to come up with that height equation?
you have constant acceleration
h" = -9.8 m/s^2
so, the speed is h' = -9.8t+Vo m/s
Vo=20, so
h' = -9.8t+20
h= -4.9t^2+20t+Ho
Ho=20, so
h = -4.9t^2+20t+30
h" = -9.8 m/s^2
so, the speed is h' = -9.8t+Vo m/s
Vo=20, so
h' = -9.8t+20
h= -4.9t^2+20t+Ho
Ho=20, so
h = -4.9t^2+20t+30
1 answer