v0=2.7 (package was rising)
x0=11
x1=0
a=-9.8
x1=x0+v1*t+(1/2)at²
0=11+(2.7)t+(1/2)(-9.8)t²
Solve for t to get t=1.8 sec.
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