hey can someone help me with the question below i will medal
I dont have any idea how to do this Arianna kicks a soccer ball off the ground and into the air with an initial velocity of 42 feet per second. Assume the starting height of the ball is 0 feet. Approximately what maximum height does the soccer ball reach? 1.3 ft 2.6 ft 26.0 ft 27.6 ft
can you help me
Hmm..there's a formula for this..
It's something like \(\sf H(t) = -16t^2 + vt + s\)
so how do you solve it?
We plug in the initial velocity(v) and the starting height(0). \(\sf H(t) = -16t^2 + vt + s\) \(\sf H(t) = -16t^2 + 42t + 0\) or \(\sf H(t) = -16t^2 + 42t\)
Now, we want to find the maximum height which is the same as the y-value of the vertex. Formula to find the x-value of a parabola: \(\sf\dfrac{-b}{2a}\) Our equation is in the form of: \(\sf ax^2 + bx + c\)
So in this case, 'a' is equal to -16, and 'b' is equal to 42.
\(\sf\dfrac{-42}{2(-16}\) Simplify that
(-16)
wouldnt it be -42/-32
Yes, now divide that.
1.3125
so what do i do now
thanks
Yes, now plug that back into the equation for 'x' or 't' in this case: \(\sf H(t) = -16t^2 + 42t\) \(\sf H(t) = -16(1.3125)^2 + 42(1.3125)\) Simplify
No, that's incorrect..
Sorry, my internet just went out for a minute there.
oh ok so what do i do wit that then
Well, first simplify \(\sf (1.3125)^2\)
1.7226563...
is the answer D
Yep, you got it.
thanks do you think that you could help me with another one in a new thread
\(\sf H(t) = -16t^2 + 42t\) \(\sf H(t) = -16(1.3125)^2 + 42(1.3125)\) Simplify the exponent: \(\sf H(t) = -16(1.72265625) + 42(1.3125)\) Multiply: \(\sf H(t) = -27.5625 + 55.125\) Add: \(\sf H(t) = 27.5625\) \(\sf H(t) \approx 27.6\)
Sure thing, I might be able to help.
ok thanks
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