A boy throws a stone upward with a velocity of 60m/s.

(a) How long will it take to reac h the maximum height (g = -10m/

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1. Answer:

   [tex]\huge\bf{{\color{maroon}{given}}}[/tex]

   [tex]u=60m/s; [/tex]

   [tex]g=-10m/s²[/tex]

   [tex]v=0[/tex]

   [tex]\textsf{(a) The time to reach maximum height is; }[/tex]

   [tex]v=u+at=u+gt[/tex]

   [tex]0=60-10t [/tex]

   [tex]t= \frac{60}{10}=6s[/tex]

   [tex]\textsf{(b) The maximum height is; }[/tex]

   [tex]v²=u²+2gs[/tex]

   [tex]- \frac{u²}{2g}= \frac{60²}{2×10}[/tex]

   [tex]=180m [/tex]

   (c) The time to reach top is equal to time taken to reach back to ground.

   Thus, time to reach the ground after reaching top is 6s.

   Or the time to reach ground after throwing is 6+6=12s.

   [tex]{\huge{\boxed{ \mathcal{\green{ Hope \: it \: \: helps }}}}}[/tex]

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