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? t 0 1 2 3 4 5 H 0 20.1 30.4 30.9 21.6 2.5 t 0 1 2 3 4 5 H 2 27 52 77 102 127 t 0 1 2 3 4 5 H 2 22.1 42.2 62.3 82.4 102.5

Answer 1

Explanation:

A ball is thrown starting at a time of 0 and a height of 2 meters. The height of the ball follows the function as follows:

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

We need to find the height of the ball from 0 to 5 seconds

At t = 0 s

`H(0)=-4.9t^2+25t+2\\\\=-4.9(0)^2+25(0)+2\\\\=2\ m`

At t = 1 s

`H(1)=-4.9t^2+25t+2\\\\=-4.9(1)^2+25(1)+2\\\\=22.1\ m`

At t = 2 s

`H(2)=-4.9t^2+25t+2\\\\=-4.9(2)^2+25(2)+2\\\\=32.4`

At t = 3 s

`H(3)=-4.9t^2+25t+2\\\\=-4.9(3)^2+25(3)+2\\\\=32.9\ m`

At t = 4 s

`H(4)=-4.9t^2+25t+2\\\\=-4.9(4)^2+25(4)+2\\\\=23.6\ m`

At t = 5 s

`H(5)=-4.9t^2+25t+2\\\\=-4.9(5)^2+25(5)+2\\\\=4.5\ m`

t = 0 1 2 3 4 5

H = 2 22.1 32.4 32.9 23.6 4.5