Question 1180996
<pre>
I think you should use the four standard equations of motion, and be careful
that upward vector quantities are positive and downward ones are negative.
The tutor above failed to observe this convention.

{{{s[F]-s[0]}}}{{{""=""}}}{{{v[0]t + expr(1/2)at^2}}}
{{{v[F]}}}{{{""=""}}}{{{v[0] + at}}}
{{{v[F]^2}}}{{{""=""}}}{{{v[0]^2 + 2a(s[F]-s[0])}}}
{{{v[av]}}}{{{""=""}}}{{{matrix(1,2,"(Average","Velocity)")}}}{{{""=""}}}{{{ (v[0]+v[F])/2}}}

</pre>A boy tosses a coin upward with a velocity of +14.7m/s. Find 

(a) the maximum height reached by the coin,<pre> 

{{{v[F]^2}}}{{{""=""}}}{{{v[0]^2 + 2a(s[F]-s[0])}}}

{{{0^2}}}{{{""=""}}}{{{(14.7)^2 + 2(-9.8)(s[F]-0.49)}}}

{{{0}}}{{{""=""}}}{{{216.09 - 19.6(s[F]-0.49)}}}

{{{-216.09}}}{{{""=""}}}{{{-19.6s[F]+9.604}}}

{{{-225.694}}}{{{""=""}}}{{{-19.6s[F]}}}

{{{(-225.694)/(-19.6)}}}{{{""=""}}}{{{s[F]}}}

{{{11.515}}}{{{""=""}}}{{{s[F]}}}

Answer:  11.515 m
</pre>(b) time of flight<pre>

{{{v[F]}}}{{{""=""}}}{{{v[0] + at}}}

{{{-14.7}}}{{{""=""}}}{{{14.7 + (-9.8)t}}}

{{{-29.4}}}{{{""=""}}}{{{-9.8t}}}

{{{(-29.4)/(-9.8)}}}{{{""=""}}}{{{t}}}

{{{3}}}{{{""=""}}}{{{t}}}

Answer:  3 seconds.
</pre>(c) velocity when the coin returns to the hand. (The boy's hand is 0.49m above
    the ground.)<pre> 
That's the velocity that's equal in magnitude but opposite in direction
velocity as the velocity when it left his hand.

Answer: -14.7 m/s
</pre>(d) Suppose the boy failed to catch the coin and the coin goes to the ground,
with what velocity will it strike the ground?<pre> 

{{{v[F]^2}}}{{{""=""}}}{{{v[0]^2 + 2a(s[F]-s[0])}}}

{{{v[F]^2}}}{{{""=""}}}{{{(14.7)^2 + 2(-9.8)(0-0.49)}}}

{{{v[F]^2}}}{{{""=""}}}{{{216.09 + 2(-9.8)(-0.49)}}}

{{{v[F]^2}}}{{{""=""}}}{{{216.09 + 2(-9.8)(-0.49)}}}

{{{v[F]^2}}}{{{""=""}}}{{{225.694}}}

{{{v[F]}}}{{{""=""}}}{{{-sqrt(225.694)}}}

We take the negative square root because the final velocity is downward.

{{{v[F]}}}{{{""=""}}}{{{-15.02311552}}}

Answer: -15.02 m/s

Edwin</pre>