Expert: Richard J. Raridon - 9/10/2004

Question
i have a problem that i didnt understand. it says to use this info in order (labeled as a *)to solve this problem with 3 subquestions.

*Suppose that an object is thrown into the air with an initial upward velocity of v 0 meters per second from a height h 0 meters above the ground. then, t seconds later, its height h(t) meters above the ground is modeled by the function h(t)= -4.9t^2 + v 0 t + h 0. (this model doesnt count for air resistance.

1) a stone is thrown with an upward velocity of 14 m/s from a cliff 30 m high.
  a) find its height above the ground t seconds later.
  b) when will the stone reach its highest elevation?
  c) when will the stone hit the ground?

Answer
Think about it this way.  You throw the rock up with a certain velocity.  Gravity is pulling it down.  So eventually, the rock stops rising and starts falling.  
v0 is the initial velocity and h0 is the initial height.
The standard equation is H(t) = h0 +v0t -(1/2)at^2
In this case, a = the acceleration due to gravity, 9.8m/s^2
a) H(t) = 30 +14t -4.9t^2
b) v = v0 -gt
At the top, the velocity is zero so 0 = 14-9.8t and
t = 1.43s
c) when the stone hits the ground, H(t) = 0
so 0 = 30 +14t -4.9t^2
solve that quadradically and you get t = 4.29s