Question

A rocket launched accelerates at 3.5m/s^2 in 5.90 secs and2.98m/s^2 in the next 5.98 secs and then experiences a free fall. What time will the rocket be in air?
Assume that the rocket is launched from the ground.

Answer 1

d1 = 0.5*3.5*5.9^2 = 60.9 m
V1=a*t = 3.5*5.9 = 20.65 m/s. = Velocity
after 3.5 s.

d2 = Vo*t + 0.5a*t^2
d2 = 20.65*5.98 + 2.98*5.98^2 = 176.8 m.
V2 = a*t = 2.98*5.98 = 17.82 m/s. =
Velocity @ 176.8 m.

d3 = (V^2-Vo^2)/2g
d3 = (0-17.82^2)/-19.6 = 16.2 m. = Free fall distance up.
t = (V-Vo)/g = (0-17.82)/-9.8 = 1.82 s.
= Time to reach max. Ht.
Tr = 5.90+5.98+1.82 = 13.7 s.
h = d1 + d2 + d3
h = 60.9 + 176.8 + 16.2 = 254 m. Above
Gnd.

h = Vo*t + 0.5g*t^2 = 254 m.
0 + 4.9t^2 = 254
t^2 = 51.8
Tf = 7.2 s. = Fall time.

T = Tr + Tf = 13.7 + 7.2 = 20.9 s. =
Time in air.

Answer 2

please explain clearly and precisely

Answer 3

d1 = 0.5*3.5*5.9^2 = 60.9 m
V1=a*t = 3.5*5.9 = 20.65 m/s. = Velocity
after 3.5 s.

d2 = Vo*t + 0.5a*t^2
d2 = 20.65*5.98 + 0.5*2.98*5.98^2 = 176.8 m.
V2 = a*t = 2.98*5.98 = 17.82 m/s. =
Velocity @ 176.8 m.

d3 = (V^2-Vo^2)/2g
d3 = (0-17.82^2)/-19.6 = 16.2 m. = Free fall distance up.
t = (V-Vo)/g = (0-17.82)/-9.8 = 1.82 s.
= Time to reach max. Ht.
Tr = 5.90+5.98+1.82 = 13.7 s.
h = d1 + d2 + d3
h = 60.9 + 176.8 + 16.2 = 254 m. Above
Gnd.

h = Vo*t + 0.5g*t^2 = 254 m.
0 + 4.9t^2 = 254
t^2 = 51.8
Tf = 7.2 s. = Fall time.

T = Tr + Tf = 13.7 + 7.2 = 20.9 s. =
Time in air.