Question in description!!!

Question

A softball is thrown upward with an initial velocity of 56 feet per second from 3 feet above the ground. The balls height in feet above the ground is modeled by h(t)=-16t²+56t+3, where t is the time in seconds after the ball is released. What is the maximum height of the ball?

Answer 1

You need to find the vertex of the parabola using the formula

 

r = -b / (2a)   ---->  x-coordinate of vertex

 

where:

a = -16

b = 56

 

Plug in these values into the formula to get your x-coordinate of the vertex.  Then once you found your x-coordinate, evaluate

 

h(r)  to get the maximum height.

 

 

Answer 2

Tiara, I do not know if you had some kinematics in a physics class.  If so, here is another way to solve this:

 

h(t) = -16t² + 56t + 3

 

This is an equation for motion under constant acceleration, in this case, acceleration of gravity.  The equation is of the form

 

h(t) = (1/2)at² + v₀t + h₀ where

 

a = acceleration of gravity = -32 ft/s² (down)

 

v₀ = initial vertical speed = 56 ft/s (up)

 

h₀ = initial height = 3 ft

 

Under constant acceleration motion, the speed is

 

v(t) = v₀ + at = 56 - 32t (t in seconds, v in ft/s)

 

Since the ball is losing speed on the way up, it will reach maximum height when its speed drops to zero, and then it will begin to fall:

 

56 - 32t = 0 ⇒ t = 56/32 s = 1.75 s

 

Substitute this value of t in h(t) to get h_max: 

 

h_max = -16(1.75)² + 56(1.75) + 3 = 52 ft