Question

a stone is dropped from a height of 45m.what will be the distance travelled by it during the last second of its motion​

Answer 1

Step-by-step explanation:

Height(h)=45 m

Intial Speed must be (u)=0

Accerlation Due to Gravity (g)=10 m/s²

Using v²=u²+2as find Final Velocity

v²=0²+2×10×45

v²=900

v=√900

v=30

Final Velocity=30 m/s

Time taken

v=u+at

30=0+10t

t=30/10

T=3

Time=3 seconds

Last Second Is 3rd Second

Use Equation of Motion

$s_n = u + \frac{1}{2} a(2n - 1) \\ \\ </p><p>$

Where is Particular time.

S_n=Distance Travelled in that exclusive time.

$s_n = 0 + \frac{1}{2} \times 10(2 \times 3 - 1) \\ \\ s_n = 5 \times 5 \\ \\ s_n = 25m</p><p></p><p>$

Distance Travelled in 3rd Second=25 m

Answer 2

Here Is Your Ans ⤵

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Ans :-

➡Distance = 25 Meter

Given :-

➡Height = 45 meter

➡Initial Speed = 0

➡Acceleration Due To Gravity = 10 m/s²

To Find :-

➡Distance during The Last Second Of Its Motion

Solution :-

Final Velocity

➡v² = u² - 2as

➡v² = 0² + 2 × 10 × 45

➡v² = 900

➡v = ✓900

➡v = 30 m/s

Time Taken

➡v = u + at

➡30 = 0 + 10t

➡t = 30 ÷ 10

➡t = 3 Sec

From Equation Of Motion

$= > s_n = u + \frac{1}{2} a(2n - 1) \\ \\ = > s_n = 0 + \frac{1}{2} \times 10(2 \times 3 - 1) \\ \\ = > s_n = 5 \times 5 \\ \\ = > s_n = 25 \: \: meter$

Hence , Distance Travelled In 3 Second Is 25 meter

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