Question

HELPPPPP!! A juggler tosses a ball into the air. The ball leaves the juggler's hand 5 feet above the ground and has an initial velocity of 30 feet per second. The juggler catches the ball when it falls back to a height of 2 feet. How long is the ball in the air? Use the model h(t)=−16t2 +v0t +h0, where v0 is the initial velocity (in feet per second), h is the height (in feet), t is the time in motion (in seconds) and h0 is the initial height (in feet).
Remember to round your answer to the nearest hundredth.

Answer 1

Answer:

t = 2.525 seconds

Step-by-step explanation:

h(t) = -16t^2 + v*t + h0  h(t) = the height above the ground after t seconds v = 40 ft/s initial velocity  h0 = 4 ft the initial height of the object (the ball)  t = the time of the motion  h(t) = -16t^2 + 40*t + 4  we need to find t such that h(t) = 3 ft  3 = -16t^2 + 40*t + 4  -16t^2 + 40*t + 4 = 3  by solving we find

t = 2.525 seconds

You can round on your own.

Answer 2

Solving the quadratic equation, it is found that the ball was in the air for 1.97 seconds.

The height of the ball, in feet, after t seconds is given by:

$h(t) = -16t^2 + v(0)t + h(0)$

In this problem:

• The initial velocity is $v(0) = 30$.
• The initial height is $h(0) = 5$.

Thus, the equation is:

$h(t) = -16t^2 + 30t + 5$

The juggler catches the ball when it falls back to a height of 2 feet, thus it stays in the air until t for which:

$h(t) = 2$

Then:

$2 = -16t^2 + 30t + 5$

$-16t^2 + 30t + 3 = 0$

$16t^2 - 30t - 3 = 0$

Which is a quadratic equation with coefficients $a = 16, b = -30,c = -3$. Then:

$\Delta = b^2 - 4ac = (-30)^2 - 4(16)(-3) = 1092$

$x_{1} = \frac{30 + \sqrt{1092}}{2(16)} = 1.97$

$x_{2} = \frac{30 - \sqrt{1092}}{2(16)} = -0.1$  

The ball stays in the air for 1.97 seconds.

A similar problem is given at https://brainly.com/question/21846193