Question

A ball is dropped from a height H from rest.The ball travels H/2 in last 1.0 s. The total time taken by the ball to hit the ground is​
A) 3.85 s
B) 3.41 s
C) 2.55 s
D) 4.65 s

Answer 1

Answer:

3.41

Explanation:

last Is, ball covers H/2distance. Let total time for falling by H height is 1 second.

In (-1) second, ball falls by height.

So.

H/2 = 1/2 * g * (t - 1) ^ 2 Now, dividing Eq. (i) by Eq. (ii), we get

H/H/2

sqrt(2) = t/(t - 1)

sqrt(2t) - sqrt(2) = t

t = (sqrt(2))/(sqrt(2) - 1)= sqrt 2 i sqrt 2 +1) ( sqrt 2-1)( sqrt 2 +1) =2 + sqrt(2) = 2 + 141

t = 3.41 s

Answer 2

Answer:

(B)The time taken by ball to hit the ground is 3.41s

Explanation:

Given that,

A ball is dropped from a height H from rest and the ball travels $\frac{H}{2}$ in last 1.0s.

We are required to find the total time taken by the ball to hit the ground.For that we shall use equation of motions for the ball along its entire journey and journey along $\frac{H}{2}$ height.

Let the time taken by the ball to cover entire height be t sec.Now for entire journey the equation of motion is-

$-H=0+\frac{1}{2} (-g)t^{2}\\= > H=\frac{1}{2}gt^{2}-- > (1)$

Similarly,for covering half of the height the equation is

$-\frac{H}{2}=0+\frac{1}{2} (-g)(t-1)^{2}\\= > H=g(t-1)^{2}-- > (2)$

Comparing equation 1 and 2,we get-

$\frac{t^{2} }{2} =(t-1)^{2}\\= > t^{2}=2(t^{2}-2t+1)= > t^{2}-4t+2=0\\t=\frac{4\pm\sqrt{4^{2}-4(1)(2)} }{2(1)} =2\pm\sqrt{2} \\t\neq2-\sqrt{2}= > t=2+\sqrt{2}=2+1.414=3.41 sec$

Therefore,The time taken by ball to hit the ground is 3.41s

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