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Zorluk: OrtaQuadratic Equations

A projectile is launched from a platform. The height h(t)h(t), in meters, of the projectile tt seconds after launch is modeled by the equation h(t)=−5t2+v0t+h0h(t) = -5t^2 + v_0 t + h_0, where v0v_0 and h0h_0 are constants. If the projectile reaches its maximum height of 8080 meters at 33 seconds after launch, what is the value of h0h_0?

  1. A
    5
  2. 35Cevap
  3. C
    80
  4. D
    125

Cevap

The value of h0h_0 is 3535, which represents the initial height of the projectile.
The maximum height of 8080 meters at 33 seconds indicates that the vertex of the quadratic function is (3,80)(3, 80). Since the lead coefficient of the t2t^2 term is −5-5, we can express the function in vertex form as h(t)=−5(t−3)2+80h(t) = -5(t - 3)^2 + 80. To find the value of h0h_0, which represents the height at t=0t = 0, we substitute 00 for tt, yielding h(0)=−5(0−3)2+80=−5(9)+80=−45+80=35h(0) = -5(0 - 3)^2 + 80 = -5(9) + 80 = -45 + 80 = 35. Thus, the initial height is 3535 meters.

Adım Adım Çözüm

1
Identify the coordinates of the vertex of the parabola from the context.
The maximum height is 8080 meters at 33 seconds, so the vertex of the function is (3,80)(3, 80).
The vertex of a downward-opening parabola represents its maximum value.
2
Write the quadratic function in vertex form using the vertex (3,80)(3, 80) and the given leading coefficient a=−5a = -5.
h(t)=−5(t−3)2+80h(t) = -5(t - 3)^2 + 80
The vertex form of a quadratic equation is h(t)=a(t−d)2+kh(t) = a(t - d)^2 + k, where (d,k)(d, k) is the vertex.
3
Evaluate the function at t=0t = 0 to find the initial height h0h_0.
h(0)=−5(0−3)2+80=−5(9)+80=−45+80=35h(0) = -5(0 - 3)^2 + 80 = -5(9) + 80 = -45 + 80 = 35
The constant term h0h_0 in standard form h(t)=−5t2+v0t+h0h(t) = -5t^2 + v_0 t + h_0 represents the height when t=0t = 0.

Anahtar Kavram

Vertex of a quadratic function and converting between vertex form and standard form.
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Konular

Quadratic Equations

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