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January 31, 2015

January 31, 2015

Posted by **shone** on Tuesday, April 2, 2013 at 6:21am.

- machanics -
**Elena**, Tuesday, April 2, 2013 at 8:25amA 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…

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