If a ball is thrown straight up from the top of a building that is 407 feet high, the position in feet
above the ground is given by the function s(t) = -16t^2 + 75t + 407 where t is the number of
seconds elapsed.
a. How high is the projectile after 3 seconds?
b. How long will it take for the ball to reach a height of 450 feet above the ground?
man, they gave you the function. Just plug in the numbers to find your answers.
(a) s(3) = -16*3^2 + 75*3 + 407
(b) solve s(t) = 450
To find the position of the ball at a particular time or height, we can substitute the given time or height value into the equation s(t) = -16t^2 + 75t + 407.
a. To find the height of the projectile after 3 seconds:
We need to substitute t = 3 into the equation s(t) = -16t^2 + 75t + 407.
s(3) = -16(3)^2 + 75(3) + 407
s(3) = -16(9) + 225 + 407
s(3) = -144 + 225 + 407
s(3) = 481
Therefore, the ball is 481 feet above the ground after 3 seconds.
b. To find how long it will take for the ball to reach a height of 450 feet:
We need to solve the equation s(t) = 450 for t.
-16t^2 + 75t + 407 = 450
-16t^2 + 75t + 407 - 450 = 0
-16t^2 + 75t - 43 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Substituting a = -16, b = 75, and c = -43 into the quadratic formula:
t = (-75 ± √(75^2 - 4(-16)(-43))) / (2(-16))
t = (-75 ± √(5625 - 2752)) / -32
t = (-75 ± √(2873)) / -32
Since we are interested in the positive time, we can disregard the negative solution:
t ≈ (-75 + √(2873)) / -32
Using a calculator, we can find the approximate value:
t ≈ 4.79
Therefore, it will take approximately 4.79 seconds for the ball to reach a height of 450 feet above the ground.