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
100 meters
100 meters
138.75 meters
138.75 meters
3.19 meters
3.19 meters
38.75 meters
To find the height of the object 2.5 seconds after it was dropped, substitute t = 2.5 seconds into the equation h(t) = -9.8t^2 + 100.
h(2.5) = -9.8(2.5)^2 + 100
h(2.5) = -9.8(6.25) + 100
h(2.5) = -61.25 + 100
h(2.5) = 38.75
So, the height of the object 2.5 seconds after it was dropped is 38.75 meters.
To find the height of the object 2.5 seconds after it was dropped, we can substitute t = 2.5 into the equation h(t) = -9.8t^2 + 100.
h(2.5) = -9.8(2.5)^2 + 100
= -9.8(6.25) + 100
= -61.25 + 100
= 38.75
Therefore, the height of the object 2.5 seconds after it was dropped is 38.75 meters.
To find the height of the object 2.5 seconds after it was dropped, we can substitute the value of t = 2.5 into the equation h(t) = -9.8t^2 + 100:
h(2.5) = -9.8(2.5)^2 + 100
Simplifying the equation, we have:
h(2.5) = -9.8(6.25) + 100
h(2.5) = -61.25 + 100
h(2.5) = 100- 61.25
h(2.5) = 38.75
Therefore, the height of the object 2.5 seconds after it was dropped is 38.75 meters. The correct response is: 38.75 meters.