The equation h(t)=−9.8t^2+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
1. 3.19 meters
2. 100 meters
3. 138.75 meters
4. 38.75 meters
To find the height of the object 2.5 seconds after it was dropped, we need to substitute t = 2.5 into the equation for h(t).
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 meters
Therefore, the height of the object 2.5 seconds after it was dropped is 38.75 meters.
The correct answer is 4. 38.75 meters.
To find the height of the object 2.5 seconds after it was dropped, we need to 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
= -9.8(6.25) + 100
= -61.25 + 100
= 38.75 meters
Therefore, the height of the object 2.5 seconds after it was dropped is 38.75 meters.
The correct answer is 4. 38.75 meters.
To find the height of the object 2.5 seconds after it was dropped, we can plug in the value of t as 2.5 in the given equation h(t) = -9.8t^2 + 100.
Let's substitute t = 2.5 into the equation:
h(2.5) = -9.8(2.5)^2 + 100
= -9.8(6.25) + 100
= -61.25 + 100
= 38.75 meters
Therefore, the height of the object 2.5 seconds after it was dropped is 38.75 meters.
So, the correct option is 4. 38.75 meters.