Question

Two particles of same mass are projected simultaneously with same speed 20m/s from the top of a tower of height 20m one is projected vertically upwards and other projected horizontally.the maximunm height achieved by centre of mass from the ground will be

Answer 1

Hope u like my process
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Coming from the second projection to first lets see

For Second projectile maximum height would be the height of from where it is thrown horizontally..

So

For second projectile,
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$\boxed{ = > maximum \: \: height = \blue{\bf \underline{20 \: m}} } \\ \\ \boxed{ = > time \: taken = \sqrt{ \frac{2h}{g} }= \blue{ \sqrt{ \frac{2 \times 20}{10} } =\bf \underline{4 \: \: sec} } }$

Now for 1st projectile,
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=> maximum height from tower
$\boxed{ = > s = \frac{ {v}^{2} - {u}^{2} }{2g} = \blue{ \bold{\frac{20 {}^{2} - {0}^{2} }{2 \times 10}} } = \underline{ \green{ \bf 20m}}}$

=>So maximum height from ground

$=>\boxed{\bf \blue{ 20 + 20} =\underline{ \orange{40 m}} }$

=> Time taken to reach the tower back

$\boxed{ = > 2t =2 \bf \frac{v - u}{g} = \blue{ 2(\frac{20 - 0}{10}) } = \green{ \bold {\underline{4 \: \: sec}}} }$
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