First slide
Projection Under uniform Acceleration
Question

A ball thrown horizontally with velocity v from the top of a tower of height h reaches the ground in t seconds. If another ball of double the mass is thrown horizontally with velocity 3v from the top of another tower of height 4h. If the first ball reaches  the ground at a horizontal distance d, the second ball reaches the ground at a horizontal distance

Moderate
Solution

R = u\sqrt {\frac{{2h}}{g}} ;\;\frac{d}{R} = \frac{v}{{3v}}\sqrt {\frac{h}{{4h}}}
R = 6d

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