The initial velocity is 90 feet per second and the path of the projectile is modeled by the parametric equations.
x = (v0cos θ)t and y = h + (v0sin θ)t - 16t2
Need to find the maximum height of the projectile
A projectile is launched from ground level at an angle of 40° with the horizontal.?
2016-05-05 12:26 am
回答 (4)
2016-05-05 3:54 am
32
2016-05-05 12:51 am
Just replace t by x/v0cosθ :
y = h + (v0sin θ)x/v0cosθ - 16(x/v0cosθ)^2 = h + x.tanθ - x^2.16/(v0cosθ)^2
which is the expression of a parabola. The negative coefficient on x^2 tells us
that this curve's vertex corresponds to a maximum of y, and the extremum of
a function y = a + bx + cx^2 is reached for x = -b/2c.
So, here, y will be maximum as x = -b/2c = tanθ.cos²θ.V0²/32 = sinθ.cosθ.V0²/32
and ymax = h + sinθ.cosθ.V0²/32.tanθ - (sinθ.cosθ.V0²/32)^2.16/(v0cosθ)^2
= h + sin²θ.V0²/32 - sin²θ.V0²/64 = h + sin²θ.V0²/64
As h = 0, θ = 40° and V0 = 90 : ymax = 52.29 feet
y = h + (v0sin θ)x/v0cosθ - 16(x/v0cosθ)^2 = h + x.tanθ - x^2.16/(v0cosθ)^2
which is the expression of a parabola. The negative coefficient on x^2 tells us
that this curve's vertex corresponds to a maximum of y, and the extremum of
a function y = a + bx + cx^2 is reached for x = -b/2c.
So, here, y will be maximum as x = -b/2c = tanθ.cos²θ.V0²/32 = sinθ.cosθ.V0²/32
and ymax = h + sinθ.cosθ.V0²/32.tanθ - (sinθ.cosθ.V0²/32)^2.16/(v0cosθ)^2
= h + sin²θ.V0²/32 - sin²θ.V0²/64 = h + sin²θ.V0²/64
As h = 0, θ = 40° and V0 = 90 : ymax = 52.29 feet
2016-05-05 12:30 am
solve for t
dy/dt = 0
hence find y
dy/dt = 0
hence find y
2016-05-05 12:29 am
maximum height occurs when dy/dt = 0 = Vo sin A - 32t. So, t = Vo sin A /32
y(Vo sin A / 32) = h + (Vo sin A) * (Vo sin A) / 32 - 16 * (Vo sin A / 32)^2
y(Vo sin A / 32) = h + (Vo sin A) * (Vo sin A) / 32 - 16 * (Vo sin A / 32)^2
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