A rocket is shot straight up from 2 m above the ground with an initial velocity of 11 m/s. When does it hit the ground?

Answer 1

Gravity decelerates the rocket at a rate of 9,8m/s/s. The equation

for height of a rocket is h(t)=.5at²+vt+y where y=initial height,

v=initial velocity and a=initial acceleration

the equation for the height of the rocket in this problem is h(t) = -4.9t² + 11t +2 in meters

set it equal to zero and solve for t using the quadratic formula

It's easy to make a mistake in calculations. Plug in t=0, 1, 2, and 3

and you have an idea where the answer should be. 3 gives a negative

height. 2 gives a positive. Somewhere in between should be the

solution where the height is zero. About 2.5 or 2.4 looks like a good estimate