Question

Which is

A ball was thrown from the top of a playground jungle gym,
1.5 m high. The ball reached a maximum height of 4.2 m when it was
3 m from where it was thrown. How far from the jungle gym was the
ball when it hit the ground?

Answer 1

The ball was 6.74 meters far from the jungle gym when it hit the ground.

Explanation:

The ball was thrown from a height of 1.5 m The ball was thrown when x = 0, therefore if the ball is 3 m away from where it was thrown, where it reached the maximum height of 4.2, we then know the vertex is (3, 4.2).

Since we have the vertex we can make an equation in vertex form.

y = a (x-3)^2 + 4.2

We then have to solve for "a".

To solve for a we need to plug in a point. The only other point we were given is (0, 1.5)

y = a(x-3)^2 + 4.2

1.5 = a(0-3)^2 + 4.2

1.5 = 9a + 4.2

1.5 -4.2 = 9a + 4.2 - 4.2

-2.7 = 9a

-2.7/9 = 9a/9

a = -0.3

Now we have a complete equation.

y = -0.3(x-3)^2 + 4.2

Since they are asking for when the ball hits the ground, we know it is when y = 0 because it is no longer in the air. When y = 0 we are looking for the zeros (x-intercepts)

To find the zeros, let's change the equation to standard form.

y = -0.3(x-3)^2 + 4.2

= -0.3 (x-3)(x-3) + 4.3

= -0.3 (x^2 -3x -3x + 9) + 4.2

= -0.3 (x^2 -6x + 9) + 4.2

now expand -0.3 into the bracket

= -0.3x^2 + 1.8x -2.7 + 4.2

= -0.3x^2 + 1.8x + 1.5

To find the zeros we should factor, however this is not factorable, therefore we must use the quadratic formula.

a = -0.3 b = 1.8 c = 1.5

X = - 1.8 + or - √(1.8)^2 - 4(-0.3)(1.5)

/2(-0.3)

= -1.8 + or - √5.04

/-0.6

X1 = -1.8 + √5.04

/-0.6

= -0.74

This zero cannot be used because it is below zero (it is inadmissible)

X2 = -1.8 - √5.04

/-0.6

= 6.74

Therefore 6.74 is the answer.

Answer 2

The ball was 2.4 m away from the jungle gym when it hit the ground.

Step-by-step explanation:

To find the distance from the jungle gym to where the ball hit the ground, we can use the concept of projectile motion. The maximum height of the ball occurs when its vertical velocity becomes zero.

Using the equations of motion, we can determine the time it takes for the ball to reach this maximum height and then use that time to find the horizontal distance traveled by the ball.

Given:
Initial height (h) = 1.5 m
Maximum height (H) = 4.2 m
Horizontal distance to the maximum height (d) = 3 m

To find the time taken to reach the maximum height, we can use the equation:

H = h + (V₀y * t) - (0.5 * g * t²)

where V₀y is the initial vertical velocity and g is the acceleration due to gravity. Rearranging the equation, we have:

t = (V₀y ± sqrt(V₀y² + 2 * g * (H - h))) / g

Using the positive value of t, we can substitute it into the equation for horizontal distance:

d = V₀x * t

where V₀x is the horizontal velocity of the ball. Since the ball is thrown horizontally, V₀x is constant and equal to the initial horizontal velocity.

Solving for d, we find that the ball was 2.4 m away from the jungle gym when it hit the ground.