The height y (in feet) of a ball thrown by a child is y=-1/14 x^2 + 2x +3 where a is the horizontal distance in feet from the point at which the ball is thrown. (a) how high is the ball when it leaves the child's hand? preview feet (b) what is the maximum height of the ball? preview feet (c) how far from the child does the ball strike the ground? preview feet get : video

Part 1) Height of ball at launch = 3 feet.

Part 2) Maximum height achieved by ball = 17 feet

Part 3) Point at which ball strikes the ground = 29.43 feet from point of launch.

Step-by-step explanation:

The height as a function of position of launch is given as

Part 1)

Since at the time the ball leaves the hand of the child the distance of the ball from point of throwing is zero thus putting x = 0 in the above equation we get

Height of throw

Thus the ball is at 3 feet high when it leaves the child's hand.

Part 2)

The maximum height the ball achieves can be found by maximizing the given function

Thus differentiating equation 'i' with respect to 'x' and subsequently equating to zero we get

Thus the maximum value of y(x) occurs at x = 14 feet thus

Part 3)

At the moment the ball strikes the ground it's height from the ground should be = 0

Thus equating y(x) = 0 we get

Taking the positive value we find the distance at which the ball strikes the ground equals 29.43 feet.

according to the triangle sum theorem, all the angles in a triangle must add up to 180 degrees. if we know one angle is already 90, we do 180-90=90. the other 2 angles must add up to 90, exactly.

step-by-step explanation:

12

step-by-step explanation:

a triangle's   full angle is 180 degrees so 180 - 156 = 24. 24/2=12