Questions: Suppose the height of a thrown object is given by h(t)=-5t^2+21t+20 meters, t seconds after it is thrown. By how much did the height change between the time it was thrown and 5 seconds after it was thrown? meters. What was the rate of change over this interval? meters/second. (If necessary, round your answer to three decimal places)

Suppose the height of a thrown object is given by h(t)=-5t^2+21t+20 meters, t seconds after it is thrown.

By how much did the height change between the time it was thrown and 5 seconds after it was thrown? meters.

What was the rate of change over this interval? meters/second. (If necessary, round your answer to three decimal places)

Transcript text: Suppose the height of a thrown object is given by $h(t)=-5 t^{2}+21 t+20$ meters, $t$ seconds after it is thrown. By how much did the height change between the time it was thrown and 5 seconds after it was thrown? $\qquad$ meters. What was the rate of change over this interval? $\square$ meters/second. (If necessary, round your answer to three decimal places)

Solution

Solution Steps

Step 1: Identify the given function and the time intervals

The height of a thrown object is given by the function $ h(t) = -5t^2 + 21t + 20 $, where $ t $ is the time in seconds after it is thrown. We need to find the height at $ t = 3 $ seconds and $ t = 5 $ seconds.

Step 2: Calculate the height at $ t = 3 $ seconds

Substitute $ t = 3 $ into the function: \[ h(3) = -5(3)^2 + 21(3) + 20 \] \[ h(3) = -5(9) + 63 + 20 \] \[ h(3) = -45 + 63 + 20 \] \[ h(3) = 38 \]

Step 3: Calculate the height at $ t = 5 $ seconds

Substitute $ t = 5 $ into the function: \[ h(5) = -5(5)^2 + 21(5) + 20 \] \[ h(5) = -5(25) + 105 + 20 \] \[ h(5) = -125 + 105 + 20 \] \[ h(5) = 0 \]

Step 4: Determine the change in height

Calculate the difference in height between $ t = 3 $ seconds and $ t = 5 $ seconds: \[ \Delta h = h(5) - h(3) \] \[ \Delta h = 0 - 38 \] \[ \Delta h = -38 \]