how many feet 16

Question

Susan throws a softball upward into the air at a speed of 32 feet per second from a 56-foot
platform. The distance upward that the ball travels is given by the function d(t)=−16t2+32t+56.
What is the maximum height of the? softball? How many seconds does it take to reach the ground after first being thrown? upward? ? (Round your answer to the nearest? tenth.)
The maximum height of the softball is ______
feet.

Answer 1

Given h(t) = -16t² + 32t + 56 = -16 * (t² - 2t) + 56

 

Complete the square: h(t) = -16 * ( t² -2t +1) + 56 + 16 = 72 - 16  * ( t - 1)²

 

From this equation, the maximum height occurs when t = 1 sec and h_max = 72 ft

 

When h(t) = 0 ft, the softball is on the ground.

 

72 - 16 * (t - 2)² = 0 => (t - 1)² = 72/16 = 18/4 = (9/4) * 2

 

∴ t - 1 =3√(2)/2, only the positive value is feasible

   t = 1 + 3√(2)/2 ≈ 3.1 sec

Answer 2

h(t) = -16t^2 + 32t + 56

 

This is an equation of the form

 

h(t) = (1/2)at^2 + v0*t + h0 where

 

a = -32 ft/s^2

v0 = 32 ft/s

h0 = 56 ft

 

The velocity under these conditions is

 

v(t) = v0 + a*t = 32 - 32t

 

Maximum height is reached when v drops to zero and the ball begins to fall.  Set v(t) = 0, solve for t, and substitute t into h(t).

 

32 - 32t = 0 ⇒ t = 1 second

 

h(1) = -16(1)^2 + 32(1) + 56 = 72 feet

 

h_max = 72 feet and is reached in 1 second

 

To find how long it takes to reach the ground (h = 0) after thrown, set h(t) = 0 and solve for t.

 

-16t^2 + 32t + 56 = 0

 

t = {-32 ± √{(32)^2 - 4(-16)(56)]} / [2*(-16)]

 

t = (-32 ± 67.88) / (-32)

 

Note that only the negative square root gives positive time.

 

t = (-32 - 67.88) / (-32) = 3.12 seconds

 

The ball reaches the ground (h = 0) 3.12 seconds after being tossed up.