Discussion
A person of 6 ft. height wants to pluck a fruit which is hanging at \(\frac{1}{4}\)th distance from the top of a 20 ft. high tree. The person standing at a distance of 28 ft. from the tree throws a stone, which hits the top of the tree. The man moves towards the tree and throws another stone again at the same angle, this time hitting the fruit. Assuming the stones travelled in a straight line from the hand till the target, what is the current distance of the person from the tree ?
Answer: B
The person is now 18 ft from the tree.
The fruit hangs a quarter of the tree's height below the top: 20 - 5 = 15 ft above the ground. The stone leaves his hand at 6 ft.
\[ \begin{aligned} \text{first throw: rise} &= 20 - 6 = 14 \text{ over } 28 \\ \text{slope} &= 14/28 = 1/2 \\ \text{second throw: rise} &= 15 - 6 = 9 \\ \text{distance} &= 9 \div \tfrac{1}{2} = 18 \text{ ft} \end{aligned} \]
The same angle means the same slope, so the distance scales with the rise. 21 ft scales the 28 ft by three-quarters without allowing for the 6 ft hand height, 15 ft is the fruit's height, and 9 ft is the rise.
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