power phase
v = 52 + a t
h = 52 t + (1/2) 2.6 t^2 = 52 t + 1.3 t^2= 202.2
solve for t with quadratic eqn
v = 2.6*t = use t to get it. This is vi for next pahse
ballistic phase
hi = 202.2
vi = from power phase
a = -9.8
v = 0 at top = vi -9.8 t
solve for t in this phase
then
h = 202.2 + vi t - 4.9 t^2
A model rocket is launched straight upward with an initial speed of 52 m/s. It acceler- ates with a constant upward acceleration of 2.6 m/s2 until its engines stop at an altitude of 202.2 m.
What is the maximum height reached by the rocket?
Answer in units of m
How long after lift-off does the rocket reach its maximum height? Answer in units of s
How long is the rocket in the air? Answer in units of s
2 answers
after calculating all that i came to a max height of 206.59m but supposedly that is not the correct answer ? did i make an error- because i worked out the problem numerous times.