A vertical distance covered for t₁=10 s of accelerated motion
is h₁=at₁²/2. The speed at this height is v₁=at₁.
The distance covered at decelerated motion during t₂=15- t₁=15-10=5 s
is h₂=v₁t₂-gt₂²/2.
H= 2000 m
H=h₁+h₂=at₁²/2 + v₁t₂ - gt₂²/2=
=at₁²/2 + at₁t₂ - gt₂²/2.
a(t₁² +2 t₁t₂) =2H+ gt₂²
If g=10 m/s²,
a={2H+ gt₂²}/(t₁² +2 t₁t₂)=
={4000+10•25}/(100+2•10•5)=21.25 m/s²
Max velocity is v₁=at₁=21.25•10=212.5 m/s=765 km/h
The aircraft covered the distance h₁=at₁²/2 =21.25•10²/2 =1062.5 m at accelerated motion and the distance h₃=4000- h₁=4000-1062.5 =2937.5 m at decelerated motion. This motion takes the time t₃.
h₃=v₁t₃-gt₃²/2.
gt₃² -2 v₁t₃ +2h₃=0
Solve for t₃
v₃= v₁-gt₃=…
But for this given data, I’ve calculated the height h₀ at which the velocity becomes zero
0=v₁ -gt => t=v₁/g=212.5/10=21.25 s
h₀=v₁t-gt²/2=212.5•21.25 - 10•21.25²/2 = 2257.8 m. =>
The max height is h₁+h₀=1062.5+2257.8 =3320.3 m => Aircraft con’t reach the height 4000 m. Anyhow, try to check me on my math…
A space aircraft is launched straight up.The aircraft motor provides a constant acceleration for 10 seconds,then the motor stops.The aircraft's altitude 15 seconds after launch is 2 km.ignore air friction .what is the acceleration,maximum speed reached in km/h and the speed(in km/h)of the aircraft as it passes through a cloud 4 km above the ground.
1 answer
1 answer