A block at rest slides down a smooth inclined plane which makes and angle $60^{o}$ with the vertical and it reaches the ground in $t_{1}$ seconds. Another block is dropped vertically from the same point and reaches the ground in $t_{2}$ seconds. What is the ratio of $t_{1} : t_{2}$ ?

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Solution

A block at rest slide down a smooth inclined plane which make an angle $60^{\circ}$ with the vertical and it reach the ground in $t_{1}$ second
Similarly, second block reach the ground in $t_{2}$ second
Time taken to reach the bottom {t_1}
$t_{1} = \sqrt{\frac{\frac{2 h}{sin θ}}{sin θ}}$
Similarly, $t_{2} = \sqrt{\frac{\frac{2 h}{sin θ}}{sin θ}}$
$t_{1} : t_{2} = 1 : 1$

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