Algebra 2 help please!?!?!??????

Question

a dolphin jumps out of the water with an initial upward velocity of 13 meters per second. The distance above the surface is given by

d=rt - 5t^2.

a.How high will the dolphin be after .7 seconds?

b.when will the dolphin be 3 meters above the surface?

c.will the dolphin ever be 10 meters above the surface?justify your answer...

please all steps and explanations would be amazingly helpful!!!!

Answer 1

1. this is not algebra, it is kalkulus

2. the formula & all quesshuns R rong

3. I thank "r" is the vertikal komponent av its speed=start speed=13 meters/sekond.

4. gravity is bout 9.8 meter/sek^2, so vertikal speed=

vstart-gt=vstart-9.81t

5. I thank that formula is for the VERTIKAL distans it go.

6. max hite it kan go is same at hite weer yu drop sumthang & it hit ground with speed=13 meter/sek.

7. speed=gt=9.81

for speed=13, t=13/9.81...1.325 sekonds.

vertikal distans=integral av speed=0.5gt^2=9.81*1.755625/2

=8.611 meters

so it will NOT get 10 meters hi.

8. apparently teech rounded g from 9.81 tu 10, so wen divide bi 2 yu get 5t^2

Answer 2

r = rate = velocity = meter/second.  The rate is the constant speed over time, and this is an essential attribute of uniform motion problems involving distance formulas.  In this problem, you are given a distance equation d = rt - 5t^2 were r = rate, t = time(seconds), && 5 is the coefficient.  If the dolphins are jumping out of the water at a velocity of 13m/s, then let r = 13.  To find out how high the dolphin will be after 0.7 || 7 / 10 seconds, you have to do substitution.

Solution(a): Substitute the time t = 0.7 || 7 / 10 into the equation for t.  You can start this by defining the equation as a function of time t, which is: f(t) = r * t - 5 * t^2, which is thus, f(.7) = 13 * (7/10) - 5(7/10)^2 = 91 / 10 - 49 / 20, which concurrently simplifies to: f(.7) = 182 - 49 / 20 = 133 / 20.

Solution(b): This one asks for the time (t) of when the dolphin will be 3 meters above the surface of water.  By using your knowledge of algebraic linear equations, you can solve for t by moving all other terms to the other side and substituting for their respective counterparts.  Thus, f(t) = distance = 3.  This implies setting up the equation with the following values: 3 = 13 * t - 5 * t^2.  By focussing on the form, this equation resembles a general quadratic equation of the form: ax^2 + bx + c, where c = 3, b = -13, && a = 5.  Thus, the quadratic equation is: 5t^2 - 13t + 3 = 0.  This quadratic equation appears to be unfactorable.  Therefore, the only way to solve for t would be to use the quadratic formula, which is of the form: t = -b +-squarerootof(b^2 - 4 * a * c) / 2*a.  By using the quadratic formula and substituting for the respective terms, you get t = 13 +-squarerootof(169 - 60) / 10 = 13 +-squarerootof(109) / 10.  

Solution(c): Same approach for this part except that this time the quadratic formula will not suffice because it will be very impossible to reckon.  The discrimant will result to squarerootof(169 - 4(50)) = squarerootof(-31))(undefined).  Thus, it will be quite impossible for the doplhin to be 10 meters above the surface except when t = 2403/310 (above 10 meters).

J.C

Answer 3

For problem A, R is 13m/s, and T is 0.7s.  Plug the values in and solve.

For B, plug 3m in for D and solve for T.  Note that there may be 2 solutions, one on the way up, and one on the way down.

For C, plug 10m in and see if there's a real solution for D.  Note that

there could be 0 (doesn't get that high), 1 (at the apex), or 2 (on the way up and on the way down) solutions.