Question

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■An Airplane accelerated down a Runway at 3.20 m/s^2 For 32.8 s .until is Finally lift off the Ground . Determine the distance traveled before takeoff.

■ A kangaroo is capable of jumping to a height of 2.62 m. Determine the takeoff speed of the kangaroo

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Answer 1

answer -

1. Given,

the initial velocity = 0 m /s.

acceleration = 3.20 m / s^2

time = 32.8 s

According to laws of motion.

s = ut + 1/2 at ^2

s = 1/2 at²

s=1/2(3.20)(32.8)²

s= 1721.344 m

the distance traveled before takeoff is 1731.3m

2. A = 9.81 M/S²

U = ?

V = 0 M/S

USING 3RD EQUATION OF UNIFORMLY ACCELERATED MOTION,

V² = U² + 2GH

0² = U² + 2×(-9.81)×2.62

-U² = -51.40

U = √51.40

U ≈ 7.17 M/S

THE TAKEOFF SPEED OF THE KANGAROO WILL BE 7.17 M/S

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Answer 2

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Here is ur Answer ⤵

1)

Using Newtons Law

S=ut +1/2 at²

===0(32.8) + 1/2 (3.2)(32.8)²

===1721.3 m

2)

Maximum height reached by Kangaroo =2.62 m

Final velocity = 0

g = -9.8

Using NLM

V² - U² = 2as

0 - u² = 2(-9.8)2.62

u===7.17 m/s

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