Use the following parametric equation:
x=(vo cos theta)t
y=h+(vo sin theta)t-16t^2
h=ft above ground
theta= angle from the horizontal.
The right field fence in a ballpark is 10 ft high and 314 ft from home plate. A baseball is hit at a point 2.5 ft above the ground. It leaves the bat at an angle of theta=40 degree, with the horizontal at the speed of 105 ft per second.
Q: Find the horizontal distance that the baseball travels.
Stacey Warren - Expert brainly.com
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x is the horizontal position
determine y for zeros; that should give you time in the air, use that t to determine the x
but then you must plug in all the given information, and then find "t" from the x equation
then plug in "t" into the y equation
i did all that but my number still comes out wrong T.T
Can someone possibly plug in the numbers and help me realize my wrong?
2.5+105sin40 t -16 t^2 = 0
t = (-105sin(40) + sqrt(105^2sin^2(40)-4(-16)(2.5))) /2(2.5)
t = .235 seconds if i entered those right
x=105cos(40)(.235) = 18.9 units of measure
i see my error, a = -16, not 2.5
shouldn't a= 16/105cos40?
t = (-105sin(40) + sqrt(105^2sin^2(40)-4(-16)(2.5))) /(2(-16))
t = 4.255 is better :)
now when x = the distance of the fence we gain a t value to test in y to see if we even clear the fence to begin with
i still think a=16/105cos40
t = 4.255 when i fix my errors
because t= x/105cos40 so when 16 times x. it would come out to be (16x/105cos40)^2
you could prolly try that route, but its quite alot of extra work
|dw:1332389066506:dw|at any rate we got 2 scenarios to test out
I’m so confused T.T
t = 314/(105cos(40)); if y is greater than 10 when t = this value, we clear the fence
parametrics just measure things independant of each other.
where did you get 314. T.T
y = 2.5 + 105sin(40)(314)/(105cos(40))-16(314/(105cos(40)))^2
it clears the fence by over 10 feet :)
314 is the distance the fence is from homeplate
so if we clear the fence, and we do, we can go with our original finding of t=4.255 into x to determine the distance traveled