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Marrrta [24]
3 years ago
9

Please help with the fourth part ​

Mathematics
1 answer:
Kryger [21]3 years ago
4 0

Answer:

ofc its p/q

Step-by-step explanation:

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\bf \displaystyle \int -32\cdot dt\implies -32t+C&#10;\\\\\\&#10;\textit{object moves from \underline{rest}, so velocity is 0 at 0secs}&#10;\\\\\\&#10;-32(0)+C=0\implies C=0\implies \boxed{v(t)=-32t}&#10;\\\\\\&#10;\textit{now to get the positional s(t)}&#10;\\\\\\&#10;\displaystyle \int -32t\cdot dt\implies -16t^2+C&#10;\\\\\\&#10;\textit{the initial \underline{position} was 400ft away at 0secs}&#10;\\\\\\&#10;-16(0)^2+C=400\implies C=400\implies \boxed{s(t)=-16t^2+400}

when will it reach the ground level? let's set s(t) = 0

\bf s(t)=-16t^2+400\implies 0=-16t^2+400\implies \cfrac{-400}{-16}=t^2&#10;\\\\\\&#10;25=t^2\implies \boxed{5=t}


part B)  check the picture below

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