Question
Two particles are projected vertically upwards with speed $20 \mathrm{~m} \mathrm{~s}^{-1}$ and $100 \mathrm{~m} \mathrm{~s}^{-1}$ respectively from the top of a tower. If their times of flight are in the ratio $1: 3$, find the height of the tower $\left(\mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2}\right)$.
Step 1
This can be written as: \[T = \frac{u}{g} + \sqrt{\frac{2h}{g} + \left(\frac{u}{g}\right)^2}\] where $u$ is the initial speed, $h$ is the height of the tower, and $g$ is the acceleration due to gravity. Show more…
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