Please solve this in 2 parts as stated... thank you!
1 answer:
Solve the equation we have that:
a. S = 3V²/ 4
b. S = 108 feet
<h3>How to solve the parameters?</h3>We have the equation to be;
V = 2√3S
Where;
- V is the speed
- S is the length
Let's make S the subject from the formula, divide both sides by 2√2S = v/ 2. Take square of both sides:
Now we need to divide both sides by three:
S = 3V²/ 4
b. If v = 12miles per hour, substituting in the equation found earlier:
S = 3 (12²)/ 4
S= 432/ 4
S = 108 feet
Thus, the formula for the length is S = 3V²/ 4
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A quadratic equation is in the form of ax²+bx+c. The time at which the height of the ball is 16 feets is 0.717 seconds and 1.221 seconds.
<h3>What is a quadratic equation?</h3>A quadratic equation is an equation whose leading coefficient is of second degree also the equation has only one unknown while it has 3 unknown numbers. It is written in the form of ax²+bx+c.
The complete question is:
A ball is thrown from an initial height of 2 feet with an initial upward velocity of 31 ft/s. The ball's height h (in feet) after 7 seconds is given by the following, h=2+31t-16t². Find all values of t for which the ball's height is 16 feet. Round your answer(s) to the nearest hundredth.
The time at which the height of the ball is 16 feet can be found by,
h = 2 + 31t - 16t²
16 = 2 + 31t - 16t²
16 - 2 - 31t + 16t² = 0
16t² - 31t + 14 = 0
t = 0.717 , 1.221
Hence, the time at which the height of the ball is 16 feets is 0.717 seconds and 1.221 seconds.
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