If a ball into the air from a height of 4ft at a time t=0. The function that models this situation is h(t)= -16^2 + 63t+4, where t is measured in seconds and h is the height in feet

Question

A) what is the height of the ball after 3 seconds

b) what is the maximum height of the ball? Round to the nearest foot

c) when will the ball hit the ground

d) what domain makes sense for the function

you must show your work on parts a,b, and c to receive credit

Answer 1

First of all there is a typo in hour equation. The function h(t) should be written as :

h(t) = -16 x t² + 63 x t + 4 So for part a) we have

a) h(3) = - 16 x 3² + 63 x 3 + 4 = 49 feet

b) The ball will reach its maximum height when the derivative of the height function equals zero:

We have:: h^'(t) = -64 x t + 63 = 0 and t = 1.96875 seconds Plugging that time into h(t) we have

h(1.96875) = -16 x 1.96875² + 63 x 1.96875 + 4 = 128.01 feet or 128 feet

c) To find out when the ball hits hits the ground set h(t) = 0 and solve for t

1. 16 x t² + 63 x t + 4 = 0 or equivalently 16 x t² - 63 x t - 4 = 0 Factoring we have

( 16 x t + 1 ) x ( t - 4 ) = 0 So t = 4 since only positive time makes sense in this case

D) the domain that makes the most sense for time is between 0 and 4 seconds

Answer 2

use h(t) = -16t² + 63t +4

a) for t=3 h(3) = -16*9 + 63*9 +4 = 49 ft

b) find the vertex of the trajectory using f(-b/2a) where a=-16 b=63 so f=63/16 = 1.97 sec and

h(1.97) =-16(1.97)² +63*1.97 +4 = -62.09 +124.11 +4 = 66 ft

c) -4 = -16t² +63t +4 and 16t²-63t -8 =0 solving for t gives t= 4.06 sec appx.