A stone is dropped from a height of 100 m,wh | vedclass

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(A) Let the two stones meet at a height $h$ from the ground after time $t$.For the first stone dropped from $100 \ m$:The height of the first stone at

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$4$

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(A) Let the two stones meet at a height $h$ from the ground after time $t$.For the first stone dropped from $100 \ m$:The height of the first stone at time $t$ is given by $y_1 = 100 - \frac{1}{2}gt^2$.For the second stone projected upwards from the ground:The height of the second stone at time $t$ is given by $y_2 = 25t - \frac{1}{2}gt^2$.Since they meet at the same height,$y_1 = y_2$:$100 - \frac{1}{2}gt^2 = 25t - \frac{1}{2}gt^2$$100 = 25t$$t = \frac{100}{25} = 4 \ s$.Thus,the stones will be at the same height after $4 \ s$.

$A$ stone is dropped from a height of $100 \ m$,while another one is projected vertically upwards from the ground with a velocity of $25 \ m/s$ at the same time. The time in seconds after which they will have the same height is (acceleration due to gravity,$g = 10 \ m/s^2$):

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