Question 451571
h = -4.9t^2 + vt + s
Given: v = 45 m/s, s = 1.3 m
To find the maximum height, we take the derivative and set it equal to zero:
dh/dt = 0 = -9.8t + v
So t = v/9.8 = 45/9.8 = 4.592
Substituting the value for t into the original equation gives:
h = -4.9(4.592)^2 + 45(4.592) + 1.3
This gives h = 104.62 m
So the rocket never reaches a height of 190 m.
The graph of the trajectory is below:
{{{graph(400,400,-10,10,-20,120,-4.9x^2+45x+1.3)}}}