Solution:
Time of ascent is \( t_a = \frac{u}{g} \). The total time of flight is \( T = 3t_a = \frac{3u}{g} \). Using \( s = uT - \frac{1}{2}gT^2 \), we get \( -60 = u\left(\frac{3u}{g}\right) - \frac{1}{2}g\left(\frac{3u}{g}\right)^2 \). Solving this gives \( u = 20\text{ m/s} \).
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