A ball is dropped from a bridge which is 160 ft above a river. How long will it take the ball to hit the water? Use the equation s(t)= -16t^2+vot+so

Question

How would you do this problem?Thanks. The o stands for sub o or knot.

Answer 1

s(t) = -16t²+ v₀t + s₀

 

Where:

• s(t) = height of the ball when it hits the water = 0 ft
• v₀ = initial velocity of the ball = 0 f/s (because the ball is dropped, not thrown down)
• s₀ = initial height of the ball when dropped =160 ft

 

     0 = -16t² + 0*t + 160

     0 = -16t² + 160

 

Solve for t:

 

      16t² = 160

      t² = 10

 

      t = √10 ≅ 3.16 seconds

Answer 2

You are seeking the time when s(t) = 0, so rewrite your equation 

0 = -16 t^2 + vot + s

 

your initial velocity vo is 0, since the verb is "dropped".

the initial height, s, is given as 160 ft.

substituting

 

0 = -16t^2 + (0)t +160   Simplify

0 = -16t^2 +160              Isolate the variable

-160 = -16t^2                 Divide both sides by -16

10 = t^2                          Take the square root of both sides

3.1623 sec = t