4 years ago

When an object has reached terminal velocity, it has stopped

Physics

1 answer:

egoroff_w [7]4 years ago

4 0

Accelerating ! is the answer

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Pls a good 6th grade science project pls I will give you a lot of points

Scrat [10]

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Explanation:

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3 years ago

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9. A football punter attempts to kick the football so that it lands on the ground 67.0 m from where it is kicked and stays in th

vichka [17]

Answer:

Angle is 55.52°

and Initial Speed is v=26.48 m/s

Explanation:

Given data

$x_{o}=0m\\ y_{o}=1.23m\\a_{oy}=a_{1y}=g=-9.8m/s^{2} \\x_{1}=67.0m\\y_{1}=0m\\t_{o}=0\\a_{ox}=m/s^{2} \\t_{1}=4.50s$

Applying the kinematics equations for motion with uniform acceleration in x and y direction

$x_{1}=x_{o}+v_{ox}t_{1}=67.0m\\0+4.50v_{o}Cos\alpha =67.0m\\v_{o}Cos\alpha =14.99\\v_{o}=14.99/Cos\alpha.....(1) \\and\\y_{1}=y_{o}+v_{oy}t_{1}+(1/2)a_{oy}t_{1}^{2} =0m\\ 1+4.50v_{o}Sin\alpha+(-9.8/2)(4.5)^{2}=0\\ v_{o}Sin\alpha=21.828.....(2)$

Put the value of v₀ from equation (1) to equation (2)

$\frac{14.99}{Cos\alpha }(Sin\alpha ) =21.828\\as\\tan\alpha =Sin\alpha /Cos\alpha \\So\\14.99tan\alpha =21.828\\tan\alpha =21.828/14.99\\\alpha =tan^{-1}(21.828/14.99) \\\alpha =55.52^{o}$