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A ball is thrown from a height of 45 meters with an initial downward velocity of 10 m/s. The ball's height h (in meters) after t seconds is given by the following. h=45-10 t- 5 t^2. How long after the ball is thrown does it hit the ground?

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opudodennis

Answer:

2.16 s

Step-by-step explanation:

When it hits the ground, height h=0 hence

[tex]0=45-10t-5t^{2}[/tex]

[tex]5t^{2}+10t-45=0[/tex]

[tex]t^{2}+2t-9=0[/tex]

Using quadratic formula

[tex]x=-1+\sqrt 10\\ x=-1-\sqrt 10[/tex]

Since time can't be negative, [tex]-1+\sqrt 10[/tex]

t=2.16 s

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