Problema Solution
A model rocket is launched with an initial velocity of 230 ft/s. The height, h, in feet, of the rocket t seconds after the launch is given by
h = −16t2 + 230t.
How many seconds after launch will the rocket be 310 ft above the ground? Round to the nearest hundredth of a second.
Answer provided by our tutors
h = −16t^2 + 230
we need to find such t, t>0 that h = 310 that is
−16t^2 + 230t = 310
by solving we find:
t1 = 12.87 s
t2 = 1.51 s
click here to see the step by step solution of the equation:
the rocket will be 310 ft above the ground after 1.51 seconds (on the way up) and after 12.87 seconds (on the way down).