A rocket launched accelerates at 3.5 m/s^2 in 5.90 secs and 2.98 m/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.
Answers
S = ut + 1/2t²
u = 0 since it begins from rest.
S₁= 0.5 × 3.5 × 5.9² = 60.9 m
V₁=a × t = 3.5 × 5.9 = 20.65 m/s.
This is velocity after 3.5 s.
S₂= ut + 0.5at²
S₂ = 20.65 × 5.98 + 2.98 × 5.98² = 176.8 m.
V₂= at = 2.98 × 5.98 = 17.82 m/s.
This is velocity at 176.8 m.
S₃ = (V²- u²)/2g = For height attained vertically
g = - 9.8 m /s²
Here it will have a negative since the body experiences deceleration.
S₃= (0-17.82²/-19.6 = 16.2 m. = Free fall distance up.
t = (V - U) /g = (0-17.82) /-9.8 = 1.82 s = Time taken to reach max height
Total time = 5.90+5.98+1.82 = 13.7 s.
Total height achieved = S₁+ S₂ + S₃
H = 60.9 + 176.8 + 16.2 = 254 m
254 m above the ground.
H = ut + 0.5gt² = 254 m.
u = 0 for a free fall.
0 + 4.9t² = 254
t²= 51.8
Tf = 7.2 s. = Fall time.
Time in air = Time to achieve maximum height + Time of fall
= 13.7 + 7.2 = 20.9 s.
Answer:
Explanation:
We will the following equation of motion:
S = ut + 1/2t²
u = 0 since it begins from rest.
S₁= 0.5 × 3.5 × 5.9² = 60.9 m
V₁=a × t = 3.5 × 5.9 = 20.65 m/s.
This is velocity after 3.5 s.
S₂= ut + 0.5at²
S₂ = 20.65 × 5.98 + 2.98 × 5.98² = 176.8 m.
V₂= at = 2.98 × 5.98 = 17.82 m/s.
This is velocity at 176.8 m.
S₃ = (V²- u²)/2g = For height attained vertically
g = - 9.8 m /s²
Here it will have a negative since the body experiences deceleration.
S₃= (0-17.82²/-19.6 = 16.2 m. = Free fall distance up.
t = (V - U) /g = (0-17.82) /-9.8 = 1.82 s = Time taken to reach max height
Total time = 5.90+5.98+1.82 = 13.7 s.
Total height achieved = S₁+ S₂ + S₃
H = 60.9 + 176.8 + 16.2 = 254 m
254 m above the ground.
H = ut + 0.5gt² = 254 m.
u = 0 for a free fall.
0 + 4.9t² = 254
t²= 51.8
Tf = 7.2 s. = Fall time.
Time in air = Time to achieve maximum height + Time of fall
= 13.7 + 7.2 = 20.9 s.