Question

A football is kicked vertically upward from a height of 2 feet with an initial speed of 50 feet per second. The formula h=2+50t-16t² describes the ball's height above the ground, h, in

feet, t seconds after it was kicked. Use this formula to find the ball's height 2 seconds after it was kicked.
The ball's height, 2 seconds after it was kicked, was feet.

Answer 1

Answer: 38 feet.

Explanation:

The equation of motion of a body thrown vertically upwards:

`\displaystyle\\\boxed {h=h_0+v_0t-(gt^2)/(2) \ \ \ \ \ (1)}`,

`h_0=2 \ feet\ \ \ \ v_0=50\ feet\ per\ second.`

Substitute in formula (1) the value t=2 c:

`h=2+50*2-16*2^2\\h=2+100-16*2*2\\h=102-64\\h=38\ feet.`