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:

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.

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:

[tex]h(t) = -16t^2 + v(0)t + h(0)[/tex]

In this problem:

Thus, the equation is:

[tex]h(t) = -16t^2 + 30t + 5[/tex]

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:

[tex]h(t) = 2[/tex]

Then:

[tex]2 = -16t^2 + 30t + 5[/tex]

[tex]-16t^2 + 30t + 3 = 0[/tex]

[tex]16t^2 - 30t - 3 = 0[/tex]

Which is a quadratic equation with coefficients [tex]a = 16, b = -30,c = -3[/tex]. Then:

[tex]\Delta = b^2 - 4ac = (-30)^2 - 4(16)(-3) = 1092[/tex]

[tex]x_{1} = \frac{30 + \sqrt{1092}}{2(16)} = 1.97[/tex]

[tex]x_{2} = \frac{30 - \sqrt{1092}}{2(16)} = -0.1[/tex]  

The ball stays in the air for 1.97 seconds.

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

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