To solve the problem, we have the height of the golf ball modeled by the equation \( h(t) = -16t^2 + 45t \).
a. After how many seconds does the golf ball reach the ground?
The golf ball reaches the ground when \( h(t) = 0 \): \[ -16t^2 + 45t = 0 \]
We can factor out \( t \) from the equation: \[ t(-16t + 45) = 0 \]
Setting each factor to zero gives:
- \( t = 0 \) (the time when the ball is hit)
- \( -16t + 45 = 0 \) \[ -16t = -45 \quad \Rightarrow \quad t = \frac{45}{16} \approx 2.8125 \text{ seconds} \]
Thus, the golf ball reaches the ground after approximately \( 2.81 \) seconds.
b. What is the maximum height the golf ball reaches before beginning to fall to the ground? After how many seconds does the ball begin falling?
The maximum height occurs at the vertex of the parabola described by the quadratic equation. The formula for the time at which the vertex occurs is given by \( t = -\frac{b}{2a} \), where \( h(t) = at^2 + bt + c \). Here, \( a = -16 \) and \( b = 45 \): \[ t = -\frac{45}{2(-16)} = \frac{45}{32} \approx 1.40625 \text{ seconds} \]
To find the maximum height, substitute this value of \( t \) back into the height equation:
\[ h\left(\frac{45}{32}\right) = -16\left(\frac{45}{32}\right)^2 + 45\left(\frac{45}{32}\right) \]
First, calculate \( \left(\frac{45}{32}\right)^2 \): \[ \left(\frac{45}{32}\right)^2 = \frac{2025}{1024} \] Now substitute back into the equation: \[ h\left(\frac{45}{32}\right) = -16 \cdot \frac{2025}{1024} + 45 \cdot \frac{45}{32} \] \[ = -\frac{32400}{1024} + \frac{2025}{32} \] Convert \( \frac{2025}{32} \) to have a common denominator of 1024: \[ = \frac{2025 \cdot 32}{32 \cdot 32} = \frac{64800}{1024} \] Now combine the two terms: \[ h\left(\frac{45}{32}\right) = -\frac{32400}{1024} + \frac{64800}{1024} = \frac{32400}{1024} = 31.640625 \]
So, the maximum height is approximately \( 31.64 \) feet.
c. What is the height of the golf ball after 1.35 seconds have passed?
To find the height after \( t = 1.35 \) seconds, substitute \( t = 1.35 \) into the height equation: \[ h(1.35) = -16(1.35)^2 + 45(1.35) \] Calculate \( (1.35)^2 \): \[ (1.35)^2 = 1.8225 \] Now substitute this into the height equation: \[ h(1.35) = -16(1.8225) + 45(1.35) \] \[ = -29.16 + 60.75 = 31.59 \]
Thus, the height of the golf ball after \( 1.35 \) seconds is approximately \( 31.59 \) feet.
Summary of answers:
- a. The golf ball reaches the ground after approximately \( 2.81 \) seconds.
- b. The maximum height is approximately \( 31.64 \) feet, reached at approximately \( 1.41 \) seconds.
- c. The height after \( 1.35 \) seconds is approximately \( 31.59 \) feet.