A stone is released from a top. Another stone is dropped from a point 15 m below the top, when the first stone has reached a point 5 m below the top. Both balls reach the ground simultaneously. Determine the height of the top

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Solution

Dear student,
$A c c o r d i n g t o q u e s t i o n H - 5 = \sqrt{2 \times 5 \times 10} t + \frac{1}{2} g t^{2} - - - 1 a n d H - 15 = \frac{1}{2} g t^{2} - - - - - 2 b y e q u a t i o n 2 i n 1 w e g e t H - 5 = \sqrt{2 \times 5 \times 10} \times \sqrt{\frac{2 (H - 15)}{g}} + H - 15 H = 20 m$
Regards

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A stone is released from a top. Another stone is dropped from a point 15 m below the top, when the first stone has reached a point 5 m below the top. Both balls reach the ground simultaneously. Determine the height of the top​

Dear student, According to question H-5=2×5×10t+12gt2---1and H-15=12gt2-----2by equation 2 in 1 we get H-5=2×5×10×2(H-15)g+H-15H=20m Regards

A stone is released from a top. Another stone is dropped from a point 15 m below the top, when the first stone has reached a point 5 m below the top. Both balls reach the ground simultaneously. Determine the height of the top