Question

A man throws a ball into the air with a velocity of 96 ft/s. Use the formula h=−16 t 2 + v 0 t to determine when the height of the ball will be 48 feet. Round to the nearest tenth.

Answer 1

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

$h=-16t^2+v_0t$

where, $v_0$ 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.

$h=-16t^2+96t$

We need to find the time at which the height of ball is 48 feet.

$48=-16t^2+96t$

$16t^2-96t+48=0$      .... (1)

Quadratic formula for $ax^2+bx+c=0$ is

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

In equation a=16, b=-96 and c=48. Using quadratic formula we get

$t=\dfrac{-(-96)\pm \sqrt{(-96)^2-4(16)(48)}}{2(16)}$

$t=\dfrac{96\pm \sqrt{9216-3072}}{32}$

$t=\dfrac{96\pm 78.384}{32}$

$t=\dfrac{96+78.384}{32}$ and $t=\dfrac{96-78.384}{32}$

$t=5.449$ and $t=0.55$

Round the answer to the nearest tenth.

$t\approx 5.4$ and $t\approx 0.6$

Therefore, the height of ball will be 48 feet at t=5.4 and t=0.6.

Answer 2

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.