Question

the height of a ball thrown into air with the initial velocity of 24ft/s from a height of 6 feet above the ground is given by the equation h= -16t^2+24t+6 where t is the time (in seconds) , that the ball has been in the air . after how many seconds is the ball at a height of 14 feet

Answer 1

Answer:

t = 0.5s and 1s

Step-by-step explanation:

We are given the equation of height of the ball as: $h= -16t^2+24t+6$

We need to find the value of t when h = 14 ft

Just substitute h = 14 and solve the quadratic equation.

$h= -16t^2+24t+6\\ \\ \Rightarrow 14=-16t^2+24t+6\\ \\ \Rightarrow 16t^2-24t-6+14=0\\ \\ \Rightarrow 16t^2-24t+8=0\\ \\ \Rightarrow 2t^2-3t+1=0\\ \\ \Rightarrow 2t^2-2t-t+1=0\\ \\ \Rightarrow 2t(t-1)-1(t-1)=0\\ \\ \Rightarrow (2t-1)(t-1)=0\\ \\ \Rightarrow 2t-1=0\ \ and\ \ t-1=0\\ \\ \Rightarrow \boxed{t=\frac{1}{2} s=0.5s,\ \ 1s}$

You can check h at time t = 1/2 s and 1s.

$h_{0.5s}= -16 \times {0.5}^2+24 \times 0.5+6\\=-16 \times 0.25+24 \times 0.5+6\\=-4+12+6\\=14\ ft\\ \\ h_{1s}= -16 \times {1}^2+24 \times 1+6\\=-16 \times 1+24 \times 1+6\\=-16+24+6\\=14\ ft$

Answer 2

answer is in attachment

thank you