Question

The function s(t) = vt + h – 0.5at2 represents the height of an object, s, in feet, above the ground in relation to the time, t, in seconds, since the object was thrown into the air with an initial velocity of v feet per second at an initial height of h feet and where a is the acceleration due to gravity (32 feet per second squared). A baseball player hits a baseball 5 feet above the ground with an initial velocity of 70 feet per second. About how long will it take the baseball to hit the ground? A) 2 seconds B) 3 seconds C) 4 seconds D) 5 seconds

Answer 1

Answer:

C) 4 seconds

Explanation:

The height of an object is given by the following formula:

$s(t)=h+vt-0.5at^2\\\\a=32ft/s$

To find the flight time you equal the height s to zero:

$0=5ft+70\frac{ft}{s}\ t-0.5(32\frac{ft}{s^2})t^2\\\\0=5ft+70\frac{ft}{s}\ t-16\frac{ft}{s^2}t^2$

Then, you have a quadratic equation of the form:

$ax^2+bx+c$

you use the quadratic formula in order to find the roots of the polynomial:

$t_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\a=-16\\\\b=70\\\\c=5\\\\t_{1,2}=\frac{-70\pm \sqrt{(70)^2-4(-16)(5)}}{2(-16)}\\\\t_{1,2}=\frac{-70\pm 72.24}{-32}\\\\t_1=4.44s\approx4s\\\\t_2=-007s$

you take the positive value of t because this has physical meaning.

Hence:

the time is C) 4 seconds

Answer 2

From the data provided, the time taken for the baseball to hit the ground is 4.0 seconds

What is the time taken for the baseball to hit the ground?

The function: s(t) = vt + h – 0.5at^2 represents the height of an object

From the data provided:

• v = 70 ft/s
• final height above ground, s = 0
• h = 5 ft
• a = 32 ft/s^2
• t = ?

Substituting the values:

0 = 70t + 5 - 0.5(32)t^2

16t^2 -+ 70t - 5 = 0

solving quadratically:

$t = \frac{ - b \sqrt{b {}^{2} - 4ac} }{2a}$

$t = \frac{ - 70 \sqrt{ {70}^{2} - 4 \times ( - 16) \times 5} }{2 \times ( - 16)}$

t = 4.4 s or - 0.07

taking only positive values for t, t= 4 seconds

Therefore, the time taken for the baseball to hit the ground is 4.0 seconds.

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