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Assume the acceleration of the object is a(t) = −32 feet per second per second. (Neglect air resistance.) A ball is thrown vertically upward from a height of 6 feet with an initial velocity of 81 feet per second. How high will the ball go? (Round your answer to two decimal places.)

Answer :

Answer:

[tex]y_{max} = 108.5 ft[/tex]

Explanation:

As we know that the position of the ball is given as

[tex]\Delta y = v_0 t + \frac{1}{2}at^2[/tex]

[tex]y - y_o = v_0t + \frac{1}{2}at^2[/tex]

now at the maximum height the differentiation of y with respect to time must be zero

[tex]\frac{dy}{dt} - 0 = v_0 + at[/tex]

now we have

[tex]0 = 81 - 32 t[/tex]

[tex]t = 2.53 s[/tex]

now for maximum height we have

[tex]y_{max} - 6 = (81)(2.53) - \frac{1}{2}(32)(2.53)^2[/tex]

[tex]y_{max} = 6 + 204.93 - 102.4[/tex]

[tex]y_{max} = 108.5 ft[/tex]

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