Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

With what velocity a ball be projected vertically so that the distance covered by it in 5th second is twice the distance it covers in its 6th second ( $g=10 \text{ m/s}^2$ )?

58.8 m/s

49 m/s

65 m/s

19.6 m/s

Step-by-Step Solution

Formula for Distance in n-th Second: The distance covered by a particle in the $n$ -th second of uniformly accelerated motion is given by $S_n = u + \frac{a}{2}(2n - 1)$ . For vertical upward motion, $a = -g = -10 \text{ m/s}^2$ .
Calculate Distances:

Distance in 5th second ( $n=5$ ): $S_5 = u - \frac{10}{2}(2 \times 5 - 1) = u - 5(9) = u - 45$
Distance in 6th second ( $n=6$ ): $S_6 = u - \frac{10}{2}(2 \times 6 - 1) = u - 5(11) = u - 55$

Apply Condition: The problem states $S_5 = 2 S_6$ . $u - 45 = 2(u - 55)$ $u - 45 = 2u - 110$ $u = 110 - 45 = 65 \text{ m/s}$
Verification of Direction: The time to reach maximum height is $t_{max} = u/g = 65/10 = 6.5 \text{ s}$ . Since both the 5th second ( $t=4$ to $5$ ) and 6th second ( $t=5$ to $6$ ) occur before $t_{max}$ , the ball is moving purely upwards during these intervals, so distance equals displacement magnitude.

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