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Question
Vikas and Vasu punt a ball into the air. The equation h=−16t2+60t represents the height of ball in feet, t seconds after it was punted for Vikas's ball. Which of the following can be Akanksha's ball height equation if her ball goes higher?
Vikas and Vasu punt a ball into the air. The equation h=−16t2+60t represents the height of ball in feet, t seconds after it was punted for Vikas's ball. Which of the following can be Akanksha's ball height equation if her ball goes higher?
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Solution
The correct option is B h=−8t(2t2−9)
Equation for Vikas's ball h=−16t2+60tTime of maximum height can be found by putting h′(t)=0h′(t)=−32t+60=0⟹t=158 secNow h′′(t)=−32<0 Hence the height is maximum at t=158
maximum height =−16(158)2+60(158)=56.25 feet
Now we need to find from the given options which ball goes higher than 56.25 feet
Option A: h=−16(t2−3t)h′(t)=−16(2t−3)=0⟹t=32h′′(t)=−32<0 Hence height is maximum.Maximum height =−16((32)2−3×32)=36<56.25
Option B: h=−8t(2t2−9)=−16t3+72th′(t)=−48t2+72=0⟹t=√1.5h′′(t)=−96t<0 at t=√1.5. Hence height is maximum.Maximum height =−16(1.5)3/2+(72√1.5)=58.787>56.25Hence option B is the correct equation.
Option C: h=−4(2t2−5)2+48h′(t)=−8(2t2−5)(4t)=0⟹t=0,t=√2.5h′(t)=−32(2t3−5t)h′′(t)=−32(6t2−5) At t=0,h′′(t)>0 Hence height is not maximum.At t=√2.5,h′′(t)=−32(10)<0 Hence height is maximum.Maximum height =−4((2×2.5)−5)+48=48<56.25
Option D: h=−4(2t2−6)2+52h′(t)=−8(2t2−6)(4t)=0⟹t=0,t=√3h′(t)=−32(2t3−6t)h′′(t)=−32(6t2−6) At t=0,h′′(t)>0 Hence height is not maximum.At t=√3,h′′(t)=−32(12)<0 Hence height is maximum.Maximum height =−4((2∗3)−6)+52=52<56.25
Equation for Vikas's ball h=−16t2+60t
Time of maximum height can be found by putting h′(t)=0
h′(t)=−32t+60=0⟹t=158 sec
Now h′′(t)=−32<0 Hence the height is maximum at t=158
maximum height =−16(158)2+60(158)=56.25 feet
Now we need to find from the given options which ball goes higher than 56.25 feet
Option A: h=−16(t2−3t)
h′(t)=−16(2t−3)=0⟹t=32
h′′(t)=−32<0 Hence height is maximum.
Maximum height =−16((32)2−3×32)=36<56.25
Option B: h=−8t(2t2−9)=−16t3+72t
h′(t)=−48t2+72=0⟹t=√1.5
h′′(t)=−96t<0 at t=√1.5. Hence height is maximum.
Maximum height =−16(1.5)3/2+(72√1.5)=58.787>56.25
Hence option B is the correct equation.
Option C: h=−4(2t2−5)2+48
h′(t)=−8(2t2−5)(4t)=0⟹t=0,t=√2.5
h′(t)=−32(2t3−5t)
h′′(t)=−32(6t2−5)
At t=0,h′′(t)>0 Hence height is not maximum.
At t=√2.5,h′′(t)=−32(10)<0 Hence height is maximum.
Maximum height =−4((2×2.5)−5)+48=48<56.25
Option D: h=−4(2t2−6)2+52
h′(t)=−8(2t2−6)(4t)=0⟹t=0,t=√3
h′(t)=−32(2t3−6t)
h′′(t)=−32(6t2−6)
At t=0,h′′(t)>0 Hence height is not maximum.
At t=√3,h′′(t)=−32(12)<0 Hence height is maximum.
Maximum height =−4((2∗3)−6)+52=52<56.25
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