A shuttle cork hitted upward from badminton racket with a velocity of 50 m / s and it reaches 3 m from the hitting point in last seconds of its upward journey. If the same shuttle cork is hitted upward with a velocity of 200 m / s then what will be the distance travelled in last second of its upward journey?

Question

A shuttle cork hitted upward from badminton racket with a velocity of 50 m / s and it reaches 3 m from the hitting point in last seconds of its upward journey. If the same shuttle cork is hitted upward with a velocity of 200 m / s then what will be the distance travelled in last second of its upward journey? (A) 111 m (B) 170 m (C) 153 m (D) 120 m

Tardigrade Admin · Accepted Answer

S = u -( a /2)(2 n -1) 3=50-(10/2)(2 n-1) 3=50-((20 n /2)-5) ⇒ 3=50-10 n+5 ⇒ 10 n=52 ⇒ n =5.2 S =200-(10/2)(2 × 5.2-1) =200-47=153

$111 m$

$170 m$

$153 m$

$120 m$

$S = u - \frac{a}{2} (2 n - 1)$
$3 = 50 - \frac{10}{2} (2 n - 1)$
$3 = 50 - (\frac{20 n}{2} - 5)$
$\Rightarrow 3 = 50 - 10 n + 5$
$\Rightarrow 10 n = 52$
$\Rightarrow n = 5.2$
$S = 200 - \frac{10}{2} (2 \times 5.2 - 1)$
$= 200 - 47 = 153$

111m

170m

153m

120m

S=u−2a​(2n−1) 3=50−210​(2n−1) 3=50−(220n​−5) ⇒3=50−10n+5 ⇒10n=52 ⇒n=5.2 S=200−210​(2×5.2−1) =200−47=153

$111 m$

$170 m$

$153 m$

$120 m$

$S = u - \frac{a}{2} (2 n - 1)$
$3 = 50 - \frac{10}{2} (2 n - 1)$
$3 = 50 - (\frac{20 n}{2} - 5)$
$\Rightarrow 3 = 50 - 10 n + 5$
$\Rightarrow 10 n = 52$
$\Rightarrow n = 5.2$
$S = 200 - \frac{10}{2} (2 \times 5.2 - 1)$
$= 200 - 47 = 153$