Question

a body is falling from height h it takes 8s to reach the ground. The time it takes to cover the first one fourth of height is​

Answer 1

t2=t12–√

t1=t22–√

t2=3–√t1

None of these

Answer :

A

Solution :

h=0+12gt21⇒t1=2hg−−−√

h2=0+12gt22⇒t22=2g×h2

⇒t2=2g×h2−−−−−−√=1srt(2)×2hg−−−√=t12–√

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Answer 2

Answer:

Provided that:

• Time = 8 seconds
• Initial velocity = 0 mps
• g = + 10 mps sq.

Don't be confused!

• Initial velocity cames as zero because body is falling from a height.

• g denotes acceleration due to gravity and we also know that it's universal value is 9.8 m/s sq. But here we are taking it as 10 m/s sq. as approx. And it is positive because the body is going downwards.

To calculate:

• The time it takes to cover the first one-fourth of height!?

Solution:

• The time it takes to cover the first one-fourth of height = 4 seconds

Using concept:

• Second equation of motion

Using formula:

• ${\small{\underline{\boxed{\pmb{\sf{s \: = ut \: + \dfrac{1}{2} \: gt^2}}}}}}$

Where, s denotes displacement or distance or height,u denotes initial velocity, t denotes time taken and g denotes acceleration due to gravity.

Required solution:

✴️ Firstly let us find out the total distance travelled by using second equation of motion, let's see how to do!

$\begin{gathered}:\implies \sf s \: = ut \: + \dfrac{1}{2} \: gt^2 \\ \\ :\implies \sf s = 0(8) + \dfrac{1}{2} \times 10(8)^{2} \\ \\ :\implies \sf s = 0 + \dfrac{1}{2} \times 10(64) \\ \\ :\implies \sf s = 0 + \dfrac{1}{2} \times 640 \\ \\ :\implies \sf s = 0 + 1 \times 320 \\ \\ :\implies \sf s = 0 + 320 \\ \\ :\implies \sf s = 320 \: m \\ \\ :\implies \sf Total \: distance = 320 \: metres\end{gathered}$

✴️ Now let's find out the first one-fourth distance travelled!

$\begin{gathered}:\implies \sf \dfrac{1}{4} \times 320 \\ \\ :\implies \sf 1 \times 80 \\ \\ :\implies \sf 1/4 \: distance \: = 80 \: m\end{gathered}$

✴️ Now again let us use second equation of motion to find out the time taken by the body to cover the first one-fourth of height! But now put s as 80 m.

$\begin{gathered}:\implies \sf s' \: = ut \: + \dfrac{1}{2} \: gt^2 \\ \\ :\implies \sf 80 = 0(t) + \dfrac{1}{2} \times 10(t)^{2} \\ \\ :\implies \sf 80 = 0 + \dfrac{1}{2} \times 10(t)^{2} \\ \\ :\implies \sf 80 = 0 + 1 \times 5t^2 \\ \\ :\implies \sf 80 = 0 + 5t^2 \\ \\ :\implies \sf 80 = 5t^2 \\ \\ :\implies \sf \dfrac{80}{5} \: = t^2 \\ \\ :\implies \sf 16 \: = t^2 \\ \\ :\implies \sf \sqrt{16} \: = t \\ \\ :\implies \sf 4 \: = t \\ \\ :\implies \sf t \: = 4 \: sec \\ \\ :\implies \sf Time \: taken \: = 4 \: sec\end{gathered}$