A stone is thrown with an initial velocity of 36ft/s from the edge of a bridge that is 49 ft above the ground. The height of
this stone above the ground t seconds after it is thrown is f(t)=-16t^2+36t+49 If a second stone is thrown from
the ground, then its height above the ground after t seconds is given by g(t)=-16t^2+v_(0)t where v_(0) is the initial velocity
of the second stone Determine the value of v_(0) such that the two stones reach the same maximum height.
When an object that is thrown upwards reaches its highest point (just before it starts to fall back to the ground), its
square will be zero. Therefore to find the maximum height of the object thrown from the bridge,use the
square equation, square