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Answer :
The rocket landed 92.45m below when it hit the ground.
When a body is dropped from a height, its initial velocity is 0. A body executing free fall moves with constant acceleration i.e., acceleration due to gravity
Given data -
- time = t = 4.3 sec
- paper rocket is dropped so u ( initial velocity ) = 0 and a = g ( acceleration due to gravity )
- Distance = s = ?
By using 2nd equation of motion:
[tex]s = ut + \frac{1}{2} at^2[/tex]
Equation of motion under gravity:
[tex]s = ut + \frac{1}{2} gt^2[/tex]
[tex]= 0 \times 4.3 \sec + \frac{1}{2} \times 10 \times (4.3)^{2}[/tex] ( assuming g = 10 m/s²)
[tex]= \frac{1}{2} \times 10 \times 18.49[/tex]
= 5 × 18.49
= 92.45 m
Thus we can conclude that the rocket landed 92.45m below when it hit the ground.
Learn more about equation of motion at https://brainly.com/question/25951773
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