Vi = 2.7 m/s
h = 11 + 2.7 t = 4.9 t^2
0 = 11 + 2.7 t - 4.9 t^2 at ground
4.9 t^2 - 2.7 t - 11 = 0
t = [ 2.7 +/- sqt (2.7^2 +44*4.9) ]/9.8
A hot air balloon is traveling vertically upward
at a constant speed of 2.7 m/s. When
it is 11 m above the ground, a package is
released from the balloon.
After it is released, for how long is the
package in the air? The acceleration of gravity
is 9.8 m/s
2
.
Answer in units of s
1 answer
1 answer