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2 years ago

6. One minute after takeoff, a rocket carrying the space shuttle into outer space reaches a speed of 447 m/s.

Physics

1 answer:

Sidana [21]2 years ago

7 0

Answer:

acceleration of the rocket is given as

$a = 7.45 m/s^2$

Explanation:

As we know that rocket starts from rest and then reach to final speed of 447 m/s after t = 1 min

so we have

$v_i = 0$

$v_f = 447 m/s$

$t = 1 min = 60 s$

so we have

$a = \frac{v_f - v_i}{t}$

$a = \frac{447 - 0}{60}$

$a = 7.45 m/s^2$

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Answer:

a) $t_{1}=3.49s$

b) $t_{2}=2.00s$

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a)Kinematics equation for the Stone, dropped:

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$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

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The ball reaches the ground, y=0, at t=t1:

$0=h-1/2*g*t_{1}^{2}$

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b)Kinematics equation for the Stone, with a initial speed of 20m/s:

$v(t)=v_{o}-g*t$

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

$y_{o}=h=60m$ initial position is bridge height

$v_{o}=-20m/s$ the stone is thrown straight down

The ball reaches the ground, y=0, at t=t1:

$0=h+v_{o}t_{2}-1/2*g*t_{2}^{2}$

$0=60-20t_{2}-1/2*9.83*t_{2}^{2}$

t2=-6.01 this solution does not have physical sense

t2=2.00

c)Kinematics equation for the Stone, with a initial speed of 20m/s with an angle of 30° above the horizontal:

$v(t)=v_{o}-g*t$

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

$y_{o}=h=60m$ initial position is bridge height

$v_{o}=20sin(30)=10m/s$ the stone is thrown with an angle of 30° above the horizontal

The ball reaches the ground, y=0, at t=t3:

$0=h+v_{o}t_{3}-1/2*g*t_{3}^{2}$

$0=60+10t_{3}-1/2*9.83*t_{3}^{2}$

t3=-2.62 this solution does not have physical sense

t3=4.66

the movement in x:

v=constant=20cos(30)m/s

x(t)=v*t

Xmax=v*t3=20cos(30)*4.66=80.71m

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