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who enters the water first? by how much? In a diving siblings showcase, 11 year old...

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mgoun | Student, Undergraduate | Valedictorian

Posted October 21, 2012 at 11:25 PM via web

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who enters the water first? by how much? 

In a diving siblings showcase, 11 year old Hannah takes off from a 24.61 foot platform with an initial velocity of 9.7 ft/sec at the same time that her brother Tyler, who is 3.5 years older, takes off from a 32.81 foot platform with an initial velocity of 4.51 ft/sec. round to the nearest hundredth.

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lfryerda | High School Teacher | (Level 2) Educator

Posted October 22, 2012 at 2:51 AM (Answer #1)

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The height function for constant acceleration is `h=-1/2 g t^2+v_0t+h_0` where `v_0` is the initial velocity and `h_0` is the initial height.  

For both children, the time to enter the water is found by setting the height to zero and solving for t.

For Hannah, we need to find the time by solving:

`-1/2(32)t^2+9.7t+24.61=0`

`-16t^2+9.7t+24.61=0`   use quadratic formula

`t={-9.7-sqrt{9.7^2+4(24.61)(16)}}/{2(-16)}`   using negative root to keep time positive

`=1.58`

For Tyler, solve the equation:

`-1/2(32) t^2+4.51t+32.81=0`   

`-16t^2+4.51t+32.81=0`   use quadratic formula

`t={-4.51-sqrt{4.51^2+4(16)(32.81)}}/{2(-16)}`

`=1.58`

Within a hundredth of a second, both children enter the water at exactly the same time of 1.58 seconds after they jump.

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