Question

algebra 2 hon
unit 6: quadratic functions
quiz lessons up to 7.5
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period:
date: 2026 - 02 - 05
unit 6&7 lessons practice problems quiz

1. kevin kicks a ball off the ground at an initial vertical velocity of 10 m/s

question: what is the maximum height reached by the ball?
you can estimate the gravitational constant as ( g = 10space m/s^2 )
so you can write the effect of gravity as: ( - 5t^2 )
you may want to consider:

1. what is the equation for the height of the ball ( h(t) )?
2. what is the height of the ball when it is on the ground?
3. when will the ball leave the ground and when will it return to hit the ground?
4. considering these 2 times, at what time ( t ) is the ball at its maximum height?
5. if you know when the ball is at it maximum height, how can you then calculate what the maximum height is?

remember also to check your answer to confirm that it is correct.
the ball was kicked from the ground
it starts at zero

Explanation:

Step1: Determine the height function

The height of an object in vertical motion is given by the equation \( h(t) = v_0t - \frac{1}{2}gt^2 \), where \( v_0 \) is the initial vertical velocity, \( g \) is the acceleration due to gravity, and \( t \) is time. Here, \( v_0 = 10 \, \text{m/s} \) and \( g = 10 \, \text{m/s}^2 \), so the height function is \( h(t) = 10t - \frac{1}{2}(10)t^2 = 10t - 5t^2 \).

Step2: Find the time at maximum height

For a quadratic function \( ax^2 + bx + c \), the vertex (which gives the maximum or minimum) occurs at \( t = -\frac{b}{2a} \). In the height function \( h(t) = -5t^2 + 10t \), \( a = -5 \) and \( b = 10 \). So, \( t = -\frac{10}{2(-5)} = -\frac{10}{-10} = 1 \, \text{second} \).

Step3: Calculate the maximum height

Substitute \( t = 1 \) into the height function \( h(t) \). \( h(1) = 10(1) - 5(1)^2 = 10 - 5 = 5 \, \text{meters} \).

Answer:

The maximum height reached by the ball is \( \boldsymbol{5} \) meters.