Question

A ball is thrown in the air from ground level. The height of the ball in meters at time t seconds is given by the function src="https://tex.z-dn.net/?f=%5Csf%20h%28t%29%3D-4.9t%5E2%2B30t." id="TexFormula1" title="\sf h(t)=-4.9t^2+30t." alt="\sf h(t)=-4.9t^2+30t." align="absmiddle" class="latex-formula">. At what times does the ball hit the ground? ( be sure to use proper units!)
please, show your work!
Thank you!

Answer 1

Answer:

The ball hits the ground when h(t) (the height) is equal to 0.

0=-4.9t^2+30t

0=t(-4.9t+30)

t=0 and t=6.1224…

So, the ball hits the ground when it is thrown (0 seconds) and after about 6.1224 seconds.

:)

Answer 2

Answer:

t = 6.15 seconds

Step-by-step explanation:

A ball is thrown upward with an initial velocity of 35 meters per second from a cliff that is 30 meters high. The height of the ball is given by the quadratic equation h=-4.9t^2+35t+30 where h is in meters and t in the time in seconds since the ball was thrown, find the time it takes the ball to hit the ground. Round you answer to the nearest tenth of a second.

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h(t) =-4.9t^2+35t+30

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When the ball hits the ground its height is zero.

So, solve -4.9t^2+35t+30 = 0

Use the quadratic formula:

t = [-35 +- sqrt(35^2-4*-4.9*30)]/(2(-4.9))

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t = [-35 +- sqrt(637)]/(-9.8)

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To get a positive solution:

t = [-35-25.24]/(-9.8)

t = 6.15 seconds

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