How high will a ball go if it is thrown vertically upward from a height of 6 feet with an initial velocity of 60 feet per second and the acceleration is -32 feet per second squred?

Question

Luke Alvarez · Accepted Answer

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Scarlett Allison · Answer

undefined To find the maximum height reached by the ball, you can use the kinematic equation:

$$ h = h_0 + v_0t + \frac{1}{2}at^2 $$

Where:
$ h $ = final height (maximum height)
$ h_0 $ = initial height (6 feet in this case)
$ v_0 $ = initial velocity (60 feet per second in this case)
$ a $ = acceleration due to gravity (-32 feet per second squared)
$ t $ = time taken to reach maximum height

At maximum height, the velocity of the ball is zero. So, we can use this fact to find the time it takes to reach the maximum height:

$$ 0 = v_0 + at $$

Solving for $ t $:

$$ t = \frac{-v_0}{a} $$

Substitute the given values:

$$ t = \frac{-60}{-32} = \frac{15}{8} \text{ seconds} $$

Now, substitute $ t $ into the equation for height:

$$ h = 6 + (60) \left( \frac{15}{8} \right) + \frac{1}{2}(-32) \left( \frac{15}{8} \right)^2 $$

$$ h = 6 + \frac{450}{8} - \frac{900}{8} $$

$$ h = 6 + \frac{450 - 900}{8} $$

$$ h = 6 - \frac{450}{8} $$

$$ h = 6 - \frac{225}{4} $$

$$ h = \frac{24 - 225}{4} $$

$$ h = \frac{-201}{4} $$

$$ h = -50.25 \text{ feet} $$

The maximum height reached by the ball is approximately 50.25 feet above the initial position.