Quadratic Functions and Projectile Motion (Medium) - SAT Physics Practice Questions

First, determine the maximum height reached by the projectile: the vertex of the parabola \(h(t) = -4.9(t - 6)^2 + 120\) is at \(t = 6\) seconds, giving a height of 120 meters. The projectile lands at \(t = 12\) seconds, so its height at this time is \(h(12) = -4.9(12 - 6)^2 + 120 = -4.9(36) + 120 = -176.4 + 120 = -56.4\) meters below the launch point, meaning it landed 56.4 meters below the launch height. The total distance traveled is twice the maximum height (up and down): \(2 \times 120 = 240\) meters.

The height of the platform can be found by evaluating the function \(h(t) = -4.9t^2 + 19.6t + 1.5\) at \(t = 0\). Substituting \(t = 0\) into the equation gives \(h(0) = -4.9(0)^2 + 19.6(0) + 1.5 = 1.5\) meters. Therefore, the height of the platform is 1.5 meters.

First, verify the height at \(t = 2\): \(h(2) = -5(2 - 4)^2 + 80 = -5(-2)^2 + 80 = -5(4) + 80 = -20 + 80 = 60\) meters, which contradicts the given height of 45 meters. However, if the problem is correct, the height of the launch pad can be found by evaluating the function at \(t = 0\). Substituting \(t = 0\) into the equation gives \(h(0) = -5(0 - 4)^2 + 80 = -5(16) + 80 = -80 + 80 = 0\) meters. Thus, the height of the launch pad is 0 meters.

To find the maximum height, we need to find the vertex of the parabola described by \(h(t) = -4.9t^2 + 19.6t + 50\). The time \(t\) at which the vertex occurs is given by \(t = -\frac{b}{2a}\), where \(a = -4.9\) and \(b = 19.6\). So, \(t = -\frac{19.6}{2(-4.9)} = 2\) seconds. Substituting \(t = 2\) into the equation gives \(h(2) = -4.9(2)^2 + 19.6(2) + 50 = -19.6 + 39.2 + 50 = 69.6\) meters.

The height of the building can be found by evaluating the function \(h(t) = -4.9(t - 3)^2 + 45\) at \(t = 0\). Substituting \(t = 0\) into the equation gives \(h(0) = -4.9(0 - 3)^2 + 45 = -4.9(9) + 45 = -44.1 + 45 = 0.9\) meters. Therefore, the height of the building is 0.9 meters.

The maximum height corresponds to the vertex of the parabola represented by the function \(h(t) = -0.5(t - 10)^2 + 50\). The vertex form of a quadratic function \(a(x - h)^2 + k\) indicates the vertex as \((h, k)\). Here, \(h = 10\) and \(k = 50\), so the drone's maximum height is 50 meters.

The initial height of the projectile can be found by evaluating the function \(h(t) = -2(t - 4)^2 + 32\) at \(t = 0\). Substituting \(t = 0\) into the equation gives \(h(0) = -2(0 - 4)^2 + 32 = -2(16) + 32 = -32 + 32 = 0\) meters.

The highest point corresponds to the vertex of the parabola represented by the function \(h(t) = -0.1(t - 5)^2 + 10\). The vertex form of a quadratic function \(a(x - h)^2 + k\) indicates the vertex as \((h, k)\). Here, \(h = 5\) and \(k = 10\), so the car reaches its highest point at \(t = 5\) seconds.

The height of the diving board can be found by evaluating the function \(h(t) = -4.9t^2 + 19.6t + 2.5\) at \(t = 0\), since the initial height corresponds to the diving board's height. Substituting \(t = 0\) into the equation gives \(h(0) = -4.9(0)^2 + 19.6(0) + 2.5 = 2.5\) meters.

The maximum altitude corresponds to the vertex of the parabola represented by the function \(h(t) = -4.9(t - 5)^2 + 120\). The vertex form of a quadratic function \(a(x - h)^2 + k\) indicates the vertex as \((h, k)\). Here, \(h = 5\) and \(k = 120\), so the maximum altitude is 120 meters.

The height of the building can be found by evaluating the function \(h(t) = -16(t - 2)^2 + 100\) at \(t = 0\), since the initial height corresponds to the building's height. Substituting \(t = 0\) into the equation gives \(h(0) = -16(0 - 2)^2 + 100 = -16(4) + 100 = -64 + 100 = 36\) meters.

The peak height corresponds to the vertex of the parabola represented by the function \(h(t) = -0.5(t - 6)^2 + 50\). The vertex form of a quadratic function \(a(x - h)^2 + k\) indicates the vertex as \((h, k)\). Here, \(h = 6\) and \(k = 50\), so the rocket reaches its peak height at \(t = 6\) seconds.

To find the time at which the object reaches its maximum height, we need to find the vertex of the parabola described by \(h(t) = -4.9t^2 + 19.6t + 1.5\). For a quadratic function \(ax^2 + bx + c\), the time \(t\) at which the vertex occurs is given by \(t = -\frac{b}{2a}\). Here, \(a = -4.9\) and \(b = 19.6\), so \(t = -\frac{19.6}{2(-4.9)} = 2\) seconds.

The deepest point corresponds to the vertex of the parabola represented by the function \(g(t) = -2(t + 4)^2 + 12\). The vertex form of a quadratic function \(a(x - h)^2 + k\) indicates the vertex as \((h, k)\). Here, \(h = -4\) and \(k = 12\), so the submarine reaches its deepest point at \(t = -4\) hours, which means 4 hours before the start of the descent.

The maximum height occurs at the vertex of the parabola represented by the function \(h(t) = -5(t - 3)^2 + 80\). The vertex form of a quadratic function \(a(x - h)^2 + k\) indicates the vertex as \((h, k)\). Here, \(h = 3\) and \(k = 80\), so the maximum height is 80 meters.