Question

A baseball is thrown across a horizontal surface and follows a path described by the equation x^(2)+25/16y^(2) -400x=0 (all dimensions are in feet). What is the highest point the ball reaches and how far from the starting point (in horizontal distance) does this happen? (What is the apex of the trajectory?)

Answer 1

want to do it by quadratic eqs or calculus?

Answer 2

calculus

Answer 3

ok let's begin
we need to find maximum value of y
so differentiate the eq wrt to x and tell me

Answer 4

I dont know. Im all lost with all this. thanks for your effort

Answer 5

Im doing parabolas and hyperbolas equations but with this one idk what to do

Answer 6

differentiating wrt to x
\[2x + \frac{ 25 }{ 16 }(2y)\frac{ dy }{ dx } - 400 = 0\]

now from this eq rearrange and get dy/dx on one side and other terms on different side
can you?

Answer 7

well let me tell you this is neither a eq of parabola or hyperbola
so check it again

Answer 8

\[2x-400=- \frac{ 25 }{ 16 } y^2\]

Answer 9

do the same thing
but we are taking horizontal distance so we need to maximize x
so differentiate the eq wrt to x
can you?

Answer 10

\[2x=400 -\frac{ 25 }{ ? }\]

Answer 11

\[2x=400 -\frac{ 25 }{ 16 } y^2\]

Answer 12

\[x=\frac{ 6400-25y^2 }{ 8} ?????\]

Answer 13

ok but i told to differentiate
so differentiating we get
\[2\frac{ dx }{ dy } = 2\frac{ 25 }{ 16 } y \]

Answer 14

i dont get it. sorry

Answer 15

fo what value of y is dx/dy maximum
when y is 0 (figure out why)

Answer 16

0

Answer 17

so maximum value of x is 200