v0=13 m/s
S=-73 m
g=-9.8 m/s²
Use
S=v0*t+(1/2)g*t²
to solve for t.
Initial kinetic energy, KE1
=(1/2)m*v0²
Initial potential energy, PE1
=mgh
where
h=73 m
m=mass of packet
Final kinetic energy, KE2
=(1/2)mV²
Equate KE1+PE1=KE2 and solve for V, the landing velocity.
A hot-air balloon is ascending at the rate of 13 m/s and is 73 m above the ground when a package is dropped over the side. (a) How long does the package take to reach the ground? (b) With what speed does it hit the ground?
1 answer
1 answer