Respuesta :

  • Initial velocity=u=0m/s
  • Acceleration=a=40m/s^2
  • Time=t=2s
  • Distance=s

Apply second equation of kinematics

[tex]\\ \tt\longmapsto s=ut+\dfrac{1}{2}at^2[/tex]

[tex]\\ \tt\longmapsto s=\dfrac{1}{2}(40)(2)^2[/tex]

[tex]\\ \tt\longmapsto s=20(4)[/tex]

[tex]\\ \tt\longmapsto s=80m[/tex]

Answer:

According to the Given Statement

Acceleration a of model rocket is + 40m/s²

Time for the rocket accelerate t is 2 s

Its initial velocity u is 0m/s ( it is at rest )

We have to calculate the height attained by rocket .

Using Kinematic Equation

  • h = ut + ½ at²

On substituting the value we obtain

➥ h = 0×2 + ½ × 40 × 2²

➥ h = 0 + 20 × 4

➥ h = 80m

So, the height attained by the rocket is 80 metres.