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asambeis [7]
3 years ago
7

Yea, gonna need some help. Thanks

Physics
1 answer:
natulia [17]3 years ago
8 0

Answer:

t = 3.48 s

Explanation:

The time for the maximum height can be calculated by taking the derivative of height function with respect to time and making it equal to zero:

h(t) = -16t^2+v_ot+h_o\\\\\frac{dh(t)}{dt}=0=-32t+v_o\\\\v_o = 32t

where,

v₀ = initial speed = 110 ft/s

Therefore,

110 = 32t\\\\t = \frac{110}{32}\\\\

<u>t = 3.48 s</u>

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Explanation:

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\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\4x^3&-2x^2y&xz^2\end{array}\right|

So we will get

\nabla  \times \vec A (\vec r )= \left( \cfrac{\partial}{\partial y}(xz^2)-\cfrac{\partial}{\partial z}(-2x^2y)\right) \hat x - \left(\cfrac{\partial}{\partial x}(xz^2)-\cfrac{\partial}{\partial z}(4x^3) \right) \hat y + \left(\cfrac{\partial}{\partial x}(-2x^2y)-\cfrac{\partial}{\partial y}(4x^3) \right) \hat z

Working with the partial derivatives we get the curl

\nabla  \times \vec A (\vec r )=0 \hat x -z^2 \hat y -4xy \hat z

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