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An object is thrown vertically upwards in such a way that it has a speed of 19.6m/s when it reaches one half its maximum altitude what are (a) its maximum altitude (b) its velocity 1 sec after is thrown and (c) its acceleration when it reaches its maximum altitude.

please answer as soon as possible!!

(a) Let maximum altitude is

*h*.

Consider motion from one half of maximum to maximum altitude.

${v}^{2}={u}^{2}-2gs\phantom{\rule{0ex}{0ex}}\Rightarrow 0={\left(19.6\right)}^{2}-2\times 9.8\times \frac{h}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow h=39.2\mathrm{m}$

(b) Now, consider motion from ground to maximum altitude.

${v}^{2}={u}^{2}-2gs\phantom{\rule{0ex}{0ex}}\Rightarrow 0={u}^{2}-2\times 9.8\times h\phantom{\rule{0ex}{0ex}}\Rightarrow u=\sqrt{2\times 9.8\times 39.2}\phantom{\rule{0ex}{0ex}}\Rightarrow u=19.6\sqrt{2}\mathrm{m}$

Velocity after 1 sec,

$v=u-gt\phantom{\rule{0ex}{0ex}}\Rightarrow v=19.6\sqrt{2}-9.8\times 1=9.8\left(2\sqrt{2}-1\right)\mathrm{m}/\mathrm{s}$

(c) Acceleration at maximum altitude = g = 9.8 m/s

^{2 }Downwards

Regards.

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