Questions: Suppose an object is thrown upward with initial velocity of 48 feet per second from a height of 120 feet. The height of the object t seconds after it is thrown is given by h(t)=-16 t^2+48 t+120 A. Find the average velocity in the first two seconds after the object is thrown. B. Find the average velocity from t=2 to t=4.

Solution Steps

We are given the height function of an object thrown upward: \[ h(t) = -16t^2 + 48t + 120 \] We need to find the average velocity over two different time intervals.

The average velocity over a time interval \([t_1, t_2]\) is given by: \[ \text{Average Velocity} = \frac{h(t_2) - h(t_1)}{t_2 - t_1} \]

For the interval \([0, 2]\):

• \(t_1 = 0\)
• \(t_2 = 2\)

First, find \(h(0)\) and \(h(2)\): \[ h(0) = -16(0)^2 + 48(0) + 120 = 120 \] \[ h(2) = -16(2)^2 + 48(2) + 120 = -64 + 96 + 120 = 152 \]

Now, calculate the average velocity: \[ \text{Average Velocity} = \frac{h(2) - h(0)}{2 - 0} = \frac{152 - 120}{2} = \frac{32}{2} = 16 \text{ feet per second} \]

For the interval \([2, 4]\):

• \(t_1 = 2\)
• \(t_2 = 4\)

First, find \(h(4)\): \[ h(4) = -16(4)^2 + 48(4) + 120 = -256 + 192 + 120 = 56 \]

Now, calculate the average velocity: \[ \text{Average Velocity} = \frac{h(4) - h(2)}{4 - 2} = \frac{56 - 152}{2} = \frac{-96}{2} = -48 \text{ feet per second} \]

Final Answer