Question

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?

Answer 1

the answer is (0,2) ball thrown from an initial height of two feet Jeremys change jar that started with a 2$ deposit

Explanation:

Answer 2

height = 2m at t = 0s

height = 22.1m at t = 1s

height = 32.4m at t = 2s

height = 32.9m at t = 3s

height = 23.6m at t = 4s

height = 4.5m at t = 5s

Explanation:

Given equation:

H(t) = -4.9t² + 25t + 2 ----------------(i)

The height of the ball is a function of time. Therefore;

(i) At the 0th second. i.e t = 0, we get the height by substituting the value of t = 0 into equation (i). i.e

H(0) = -4.9(0)² + 25(0) + 2

H(0) = 2

∴ At t = 0, the height is 2 meters. This is also obvious in the first statement of the question.

(ii) 1st second. i.e t = 1, we get the height by substituting the value of t = 1 into equation (i). i.e

H(1) = -4.9(1)² + 25(1) + 2

H(1) = 22.1

∴ At t = 1, the height is 22.1 meters.

(iii) 2nd second. i.e t = 2, we get the height by substituting the value of t = 2 into equation (i). i.e

H(2) = -4.9(2)² + 25(2) + 2

H(2) = 32.4

∴ At t = 2, the height is 32.4 meters.

(iv) 3rd second. i.e t = 3, we get the height by substituting the value of t = 3 into equation (i). i.e

H(3) = -4.9(3)² + 25(3) + 2

H(3) = 32.9

∴ At t = 3, the height is 32.9 meters.

(v) 4th second. i.e t = 4, we get the height by substituting the value of t = 4 into equation (i). i.e

H(4) = -4.9(4)² + 25(4) + 2

H(4) = 23.6

∴ At t = 4, the height is 23.6 meters.

(vi) 5th second. i.e t = 5, we get the height by substituting the value of t = 5 into equation (i). i.e

H(5) = -4.9(5)² + 25(5) + 2

H(5) = 4.5

∴ At t = 5, the height is 4.5 meters.