The equation h(t)=−9.8t2+100 represents the relationship of the height, in meters, over time, in seconds, of an object dropped from the height of 100 meters. What is the height of the object 2.5 seconds after it was dropped?(1 point) Responses 138.75 meters 138.75 meters 3.19 meters 3.19 meters 38.75 meters 38.75 meters 100 meters

1 answer

To find the height of the object 2.5 seconds after it was dropped using the equation \( h(t) = -9.8t^2 + 100 \), we substitute \( t = 2.5 \) into the equation.

\[ h(2.5) = -9.8(2.5)^2 + 100 \]

First, we calculate \( (2.5)^2 \): \[ (2.5)^2 = 6.25 \]

Now substitute this back into the equation: \[ h(2.5) = -9.8 \times 6.25 + 100 \] \[ h(2.5) = -61.25 + 100 \] \[ h(2.5) = 38.75 \]

Thus, the height of the object 2.5 seconds after it was dropped is 38.75 meters.