A ball is thrown from the height of 35 meters with an initial downward velocity of 5m/s. The ball's height h (in meters) after t
1 answer:
You find this value by factoring. Factoring finds the zeros of the function. When the ball hits the ground, it has 0 height. Therefore, when the function goes through the x axis, it has a y value of 0. That means that when y = 0, the ball has hit the ground. Use the quadratic formula to factor 
. Notice I rearranged the terms. When you factor that you get zeros of -3.19 and 2.19. The two things in math that will never EVER be negative are time and distance/length. Therefore, that tells us that the ball hits the ground 2.19 seconds after it was thrown.
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The answer is
G = [m1(veca1a2) + m2 (veca3a4) + m3veca5a6)] / (m1 + m2 + m3)
G= m1x5 + m2 x 4sqrt (2) + m3x 3sqrt (2) / 16 = 35 +4 x 3.7x sqrt (2) + 5.3 x3 4sqrt (2) / 16 =35 + 36<span> sqrt (2) / 16
G = 53.09
LOOK AT THE IMAGE
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G = [m1(veca1a2) + m2 (veca3a4) + m3veca5a6)] / (m1 + m2 + m3)
G= m1x5 + m2 x 4sqrt (2) + m3x 3sqrt (2) / 16 = 35 +4 x 3.7x sqrt (2) + 5.3 x3 4sqrt (2) / 16 =35 + 36<span> sqrt (2) / 16
G = 53.09
LOOK AT THE IMAGE
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Answer:
An equation is
Step-by-step explanation:
Let be the measure of the unknown acute angle.
According to angle sum property of a triangle, sum of all the angles of a triangle is equal to 180°.
One of the acute angles of a right triangle is equal to 52°.
Also, the triangle is a right angled triangle, one of the angle must be equal to 90°.
So, an equation is