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.

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step-by-step explanation:

answer: option a.

step-by-step explanation:

you know that the quadrilateral opqr ( a four sided figure) is inscribed in circle. this means that all its vertices are on the circle.

then, by definition, the opposite angles of the quadrilateral are supplementary, this means that the sum of their measures is 180 degrees.

you can observe in the figure that the angles ∠opq   and the angle ∠orq   are opposite, therefore:

then, can be used to d to calculate the measure of ∠opq