Question

A model rocket rises from rest with constant acceleration of 106m/s2.what is the rocket speed at a height of 3.2m

Answer 1

We can solve the problem by using the following equation:

$v_f ^2 -v_i^2 =2aS$

where

vf is the final speed of the rocket

vi=0 is the initial speed of the rocket (it starts from rest)

$a=106 m/s^2$ is the rocket's acceleration

S=3.2 m is the distance covered by the rocket

Re-arranging the equation, we find the velocity of the rocket after 3.2 m:

$v_f = \sqrt{2aS}=\sqrt{2(106 m/s^2)(3.2 m)}=26.0 m/s$

Answer 2

At a height of 3.2 meters, the final speed of the model rocket is 26.05 m/s.

Given the following data:

• Initial speed = 0 m/s (since the model rocket accelerated from rest).

• Acceleration = 106 $m/s^2$

• Distance (height) = 3.2 meters.

To find the final speed of the model rocket, we would use the third equation of motion;

$V^2 = U^2 + 2aS$

Where:

• V is the final speed.

• U is the initial speed.

• a is the acceleration.

• S is the displacement or distance covered.

Substituting the given parameters into the formula, we have;

$V^2 = 0^2 + 2(106)(3.2)\\\\V^2 = 0 + 678.4\\\\V = \sqrt{678.4}$

Final velocity, V = 26.05 m/s

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