Two balls are dropped from the same point after an interval of $1$ second. Their separation $3$ second after the release of the second ball. ( $g = 10 m / s$ $^{2}$ )

25 m

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30 m

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35 m

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40 m

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Solution

The correct option is A $35 m$

distance travelled by ball in time $t$ is $\frac{1}{2} g t^{2}$

separation : $s = \frac{1}{2} g 4^{2} - \frac{1}{2} g 3^{2} = 5 \times 16 - 5 \times 7 = 35$

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Two balls are dropped from the same point after an interval of 1 second. Their separation 3 second after the release of the second ball. (g=10m/s 2)

The correct option is A 35m distance travelled by ball in time t is 12gt2 separation : s=12g42−12g32=5×16−5×7=35

Two balls are dropped from the same point after an interval of $1$ second. Their separation $3$ second after the release of the second ball. ( $g = 10 m / s$ $^{2}$ )

The correct option is A $35 m$

distance travelled by ball in time $t$ is $\frac{1}{2} g t^{2}$

separation : $s = \frac{1}{2} g 4^{2} - \frac{1}{2} g 3^{2} = 5 \times 16 - 5 \times 7 = 35$