Question

a pumpkin was tossed in the air. the height of the pumpkin in feet can be represented by $h(t) = -16t^2 + 60t + 7$, for $t$ seconds. at approximately, what time does the pumpkin reach the ground?
a 1.875 seconds
b 3.8633 seconds
c 4.4855 seconds
d 7 seconds
e 0.1133 seconds

Explanation:

Step1: Set height to zero

When the pumpkin reaches the ground, \( h(t) = 0 \). So we solve the equation \( -16t^2 + 60t + 7 = 0 \).

Step2: Use quadratic formula

For a quadratic equation \( ax^2 + bx + c = 0 \), the solutions are \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = -16 \), \( b = 60 \), \( c = 7 \).

First, calculate the discriminant: \( \Delta = b^2 - 4ac = 60^2 - 4(-16)(7) = 3600 + 448 = 4048 \).

Then, \( t = \frac{-60 \pm \sqrt{4048}}{2(-16)} = \frac{-60 \pm \sqrt{4048}}{-32} \).

We take the positive root (since time can't be negative). \( \sqrt{4048} \approx 63.62 \).

So \( t = \frac{-60 - 63.62}{-32} \) (we take the minus sign in the numerator because the plus sign would give a negative time) \( = \frac{-123.62}{-32} \approx 3.863 \) seconds.

Answer:

B. 3.8633 seconds