Question

A baseball is hit in the air. Its height above the ground is described by the function
Ht=-16t^2+45t+5, where H(t) represents the height in feet of the ball t seconds after it is hit. To the nearest hundredth of a second, for how much time will the ball be in the air? Find algebraically.
help guys plz i suck at math, also step by step is needed Branliest is the reward like always

Answer 1

Answer:

2.9195

Step-by-step explanation:

To find the time we need to equate the function the zero and solve for t:

${ - 16t}^{2} + 45t + 5 = 0$

Using the general formula we have that:

$t = \frac{ - 45 + \sqrt{ {45}^{2} - 4 ( - 16)(5)} }{2( - 16)} \\ = \frac{ - 45 + \sqrt{2345} }{ - 32}$

then:

$t = \frac{ - 45 + \sqrt{2345 } }{ - 32} = - 0.107 \\ or \\ t = \frac{ - 45 - \sqrt{2345} }{ - 32} = 2.9195$

Since the time has to be positive we conclude that the ball is in the air 2.9195 seconds.