Use a(t) = -9.8 meters per second per second as the acceleration due to gravity. (Neglect air resistance.) A baseball is thrown upward from a height of 2 meter

Question

Use a(t) = -9.8 meters per second per second as the acceleration due to gravity. (Neglect air resistance.)

A baseball is thrown upward from a height of 2 meters with an initial velocity of 10 meters per second. Determine its maximum height. (Round your answer to two decimal places.)

Answer 1

h(t) = (1/2)at² + v₀t + h₀

• h(t) = height at time t
• a = acceleration due to gravity = -9.8 m/s²
• t = time
• v₀ = initial velocity = 10 m/s
• h₀ = initial height = 2 m

 

h(t) = (1/2)at² + v₀t + h₀

h(t) = -4.9t² + 10t + 2

 

Take the derivative of h(t) wrt t, set it to zero, and solve for t.  You will get 2 answers.  The positive one is the one to use (can't go back in time just yet).  That is the time at which the ball reached its maximum height.  Plug that value of t into the h(t) equation to get the max height.

 

You can also solve it algebraically by converting the h(t) equation to vertex form, h(t) = a(t-h)²+k, by completing the square on h(t).  k will be the max height.