Amy throws a softball in the air with an initial velocity of 50 feet per second. She throws the softball from a height of 6 feet. The height of the softball, h, is modeled by the function h (t) = -16t^2 + 50t + 6, where t is the time in seconds. How high is the softball after 1.5 seconds?
well, just substitute 1.5 for every t.
Amy throws a softball in the air with an initial velocity of 50 feet per second. She throws the softball from a height of 6 feet. The height of the softball, h, is modeled by the function h (t) = -16t2 + 50t + 6, where t is the time in seconds. How high is the softball after 1.5 seconds?
To find the height of the softball after 1.5 seconds, we need to substitute t = 1.5 into the function h(t) = -16t^2 + 50t + 6.
Step 1: Substitute t = 1.5 into the equation.
h(1.5) = -16(1.5)^2 + 50(1.5) + 6
Step 2: Simplify the equation.
h(1.5) = -16(2.25) + 75 + 6
h(1.5) = -36 + 75 + 6
h(1.5) = 45
The height of the softball after 1.5 seconds is 45 feet.
To find the height of the softball after 1.5 seconds, we can substitute t = 1.5 into the equation h(t) = -16t^2 + 50t + 6.
h(1.5) = -16(1.5)^2 + 50(1.5) + 6
Now, let's simplify the equation:
h(1.5) = -16(2.25) + 75 + 6
h(1.5) = -36 + 75 + 6
h(1.5) = 45 + 6
h(1.5) = 51
Therefore, the height of the softball after 1.5 seconds is 51 feet.