With what velocity a ball be projected vertically so that the distance covered by it in 5th second is twice th

Question

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 ($g=10 	ext{ m/s}^2$)?

Accepted Answer

1. **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, $a = -g = -10 	ext{ m/s}^2$.
2. **Calculate Distances:**
   - Distance in 5th second ($n=5$): 
     $$S_5 = u - \frac{10}{2}(2 	imes 5 - 1) = u - 5(9) = u - 45$$
   - Distance in 6th second ($n=6$):
     $$S_6 = u - \frac{10}{2}(2 	imes 6 - 1) = u - 5(11) = u - 55$$
3. **Apply Condition:** The problem states $S_5 = 2 S_6$.
   $$u - 45 = 2(u - 55)$$
   $$u - 45 = 2u - 110$$
   $$u = 110 - 45 = 65 	ext{ m/s}$$
4. **Verification of Direction:** The time to reach maximum height is $t_{max} = u/g = 65/10 = 6.5 	ext{ s}$. 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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