Asked by Rebekha
A rookie quarterback throws a football with an initial upward velocity component of 15.4m/s and a horizontal velocity component of 19.5m/s . Ignore air resistance.
What is the equation for this graph and how do I find it?
What is the equation for this graph and how do I find it?
Answers
Answered by
Damon
Horizontal, no force, u =constant
so
x = 19.5 t so t = x/19.5
Vertical, f = -9.81 m/s^2 * mass = m a
so
a = acceleration = - 9.81
so
a = -9.81
v = Vi - 9.81 t
v = 15.4 - 9.81 t
then
y = initial height + 15.4 t - 4.9 t^2
call initial height of quarterback hand Hi
y = Hi + 15.4 t - 4.9 t^2
but we know t = 19.5/x
so
y = Hi + 15.4 (x/19.5) - 4.9 (x/19.5)^2
that is a parabola, multiply it out
so
x = 19.5 t so t = x/19.5
Vertical, f = -9.81 m/s^2 * mass = m a
so
a = acceleration = - 9.81
so
a = -9.81
v = Vi - 9.81 t
v = 15.4 - 9.81 t
then
y = initial height + 15.4 t - 4.9 t^2
call initial height of quarterback hand Hi
y = Hi + 15.4 t - 4.9 t^2
but we know t = 19.5/x
so
y = Hi + 15.4 (x/19.5) - 4.9 (x/19.5)^2
that is a parabola, multiply it out
Answered by
Damon
but we know t = 19.5/x
should be t = x/19.5 of course
should be t = x/19.5 of course
Answered by
Steve
wikipedia has a nice article on <u>trajectory</u>. If you scroll town a bit, you come to the equations in terms of θ, x and t.
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