Chan is throwing a ball from the ground at an initial velocity of 22.1 meters per seconds. He threw the ball so high that on it's way down it got stuck in a tree 10 meters high. How long was the ball in the air before it got stuck in the tree? a.4 sec
b.4.51 sec
c.24.9 meters
d.2.25 sec

The ball's height y at time t is

y = (22.1 m/s) t - 1/2 gt ²

where g = 9.80 m/s² is the magnitude of the acceleration due to gravity.

Solve for t when y = 10 m:

10 m = (22.1 m/s) t - 1/2 gt ²

→   t = 0.510 s or t = 4.00 s

The first time refers to when the ball first reaches 10 m on its way up, so t = 4.00 s and the answer is A.

- 2 + 10x³ - 21x² + 19x - 4

step-by-step explanation:

each term in the second factor is multiplied by each term in the first factor

x²(- 2x² + 4x - 1) - 3x(- 2x² + 4x - 1) + 4(- 2x² + 4x - 1) ← distribute

= - 2 + 4x³ - x² + 6x³ - 12x² + 3x -8x² + 16x - 4

collect like terms

= - 2 + 10x³ - 21x² + 19x - 4 ← in standard form

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step-by-step explanati∵∵∴∴on: