Answer:
Assuming that v0 represents the initial velocity, then v0 = 96 ft/s. Substitute this value into h(t) and set the equation equal to 48. Then, write this quadratic equation in standard form, which is ax² + bx+ c = 0, where a, b, and c are constants. Either use factoring or the Quadratic Function to solve the equation for t. Remember that t must be positive, because it represents a unit of time.
Answer:
t=5.4 and t=0.6.
Step-by-step explanation:
The height of the ball (in feet) after time t is defined by the function
[tex]h=-16t^2+v_0t[/tex]
where, [tex]v_0[/tex] is initial velocity.
It is given that a man throws a ball into the air with a velocity of 96 ft/s.
Substitute v=96 in the above function.
[tex]h=-16t^2+96t[/tex]
We need to find the time at which the height of ball is 48 feet.
[tex]48=-16t^2+96t[/tex]
[tex]16t^2-96t+48=0[/tex] .... (1)
Quadratic formula for [tex]ax^2+bx+c=0[/tex] is
[tex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
In equation a=16, b=-96 and c=48. Using quadratic formula we get
[tex]t=\dfrac{-(-96)\pm \sqrt{(-96)^2-4(16)(48)}}{2(16)}[/tex]
[tex]t=\dfrac{96\pm \sqrt{9216-3072}}{32}[/tex]
[tex]t=\dfrac{96\pm 78.384}{32}[/tex]
[tex]t=\dfrac{96+78.384}{32}[/tex] and [tex]t=\dfrac{96-78.384}{32}[/tex]
[tex]t=5.449[/tex] and [tex]t=0.55[/tex]
Round the answer to the nearest tenth.
[tex]t\approx 5.4[/tex] and [tex]t\approx 0.6[/tex]
Therefore, the height of ball will be 48 feet at t=5.4 and t=0.6.
