A ball is thrown starting at a time of 0 and a height of 2 meters. The height of the ball follows the function H(t)=−4.9t2+25t+2. What is the height of the ball at each second from 0 to 5? A:
t012345
H020.130.430.921.62.5'

B:
t012345
H2275277102127

C:
t012345
H222.142.262.382.4102.5

D:
t012345
H222.132.432.923.64.5

H(t) = -4.9t^2 + 25t + 2

Step-by-step explanation:

Height is a function of time

plug in 0, 1,2,3,4, and 5 to find the height at 0, 1,2,3,4, and 5 seconds.

at t = 0

H(0) = -4.9(0)^2+25(0) + 2 = 2 meters

Repeat for T=1,2,3,4 and 5

Notice the ball peaks around t = 3 seconds and starts to descend.

H(1) = (-4.9)(1^2) + 25(1) + 2 = 22.1 meters

H(2) = (-4.9)(2^2)+25(2)+2=32.4 meters

H(3) = 32.9 meters

H(4) = 23.6 meters

H(5) = 4.5 meters

step-by-step explanation:

5×5×5+2(8)÷2(2)+3(3)

15+16÷4+9

31÷13

do 31÷13 and u will get ur answer

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