Respuesta :

You're given the typical position function H=-16t2+Vt+S

You're also given that V=20

So the only thing to figure out is what is S? Well, the frog essentially leaps from the top of the water since he is on a lilly pad so his initial height is 0 since he started from the ground.

Your position polynomial would be H= -16t2+20t+0

When the frog hits the water you would simply set H equal to 0 since he will be at the ground when this happens.

So 0= -16t2+20t → 0=(-16t+20)t

Sowe know that we have the following:

t=0s
16t=20 → t=5/4s



possibly. let me know?

Answer:

t = 0, and t = 4/5

Step-by-step explanation:

1. Write the equation in standard form:

-75t^2+60t=0

2. Solve with the Quadratic formula:

[tex]t_{1,2} = \frac{-60 +- \sqrt{60^2 - 4(-75)}0}{2(-75)} \\\\t_{1,2} = \frac{-60+-60}{2(-75)} \\\\3. Separate the fractions:\\t_{1} = \frac{-60+60}{2(-75)} , t_{2} = \frac{-60-60}{2(-75)} \\\\4. Solve for each:\\t_{1} = \frac{0}{-150} = 0\\t_{2} = \frac{-120}{-150} = 4/5[/tex]

hope this helps!

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