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ddd [48]
3 years ago
8

A rocket moves upward, starting from rest with an acceleration of 25.4 m/s^2 for 3.39 s. It runs out of fuel at the end of the 3

.39 s, but does not stop. How high does it rise above the ground? Answer in units of m.
Physics
1 answer:
kolbaska11 [484]3 years ago
5 0

Explanation:

Initial speed of the rocket, u = 0

Acceleration of the rocket, a=25.4\ m/s^2

Time taken, t = 3.39 s

Let v is the final velocity of the rocket when it runs out of fuels. Using the equation of kinematics as :

v=u+at

v=25.4\times 3.39=86.10\ m/s    

Let x is the initial position of the rocket. Using third equation of kinematics as :

v^2=u^2+2ax_o

x_o=\dfrac{v^2}{2a}

x_o=\dfrac{86.10^2}{2\times 25.4}=145.92\ m  

Let x_o is the position at the maximum height. Again using equation of motion as :

v^2-u^2=2a(x-x_o)

Now a=-g and v and u will interchange

u^2=2g(x-x_o)

x=x_o+\dfrac{u^2}{2g}

x=145.92+\dfrac{(86.10)^2}{2\times 9.8}

x = 524.14 meters

Hence, this is the required solution.

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