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nirvana33 [79]
4 years ago
13

A place kicker must kick a football from a point 36.0 m (about 40 yards) from the goal. half the crowd hopes the ball will clear

the crossbar, which is 3.05 m high. when kicked, the ball leaves the ground with a speed of 23.4 m/s at an angle of 50.0° to the horizontal. (a) by how much does the ball clear or fall short of clearing the crossbar
Physics
1 answer:
trapecia [35]4 years ago
4 0
<span>The ball clears by 11.79 meters Let's first determine the horizontal and vertical velocities of the ball. h = cos(50.0)*23.4 m/s = 0.642788 * 23.4 m/s = 15.04 m/s v = sin(50.0)*23.4 m/s = 0.766044 * 23.4 m/s = 17.93 m/s Now determine how many seconds it will take for the ball to get to the goal. t = 36.0 m / 15.04 m/s = 2.394 s The height the ball will be at time T is h = vT - 1/2 A T^2 where h = height of ball v = initial vertical velocity T = time A = acceleration due to gravity So plugging into the formula the known values h = vT - 1/2 A T^2 h = 17.93 m/s * 2.394 s - 1/2 9.8 m/s^2 (2.394 s)^2 h = 42.92 m - 4.9 m/s^2 * 5.731 s^2 h = 42.92 m - 28.0819 m h = 14.84 m Since 14.84 m is well above the crossbar's height of 3.05 m, the ball clears. It clears by 14.84 - 3.05 = 11.79 m</span>
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So, 500 A/sec.

What is Solensid?

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To learn more about Solensid visit:

brainly.com/question/15061394

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What is the net work doneon the object over the distance shown?
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A)F_0d

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If you graph the force on an object as a function of the position of that object, then the area under the curve will equal the work done on that object, so we need to find the area under the function to find the work

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Area:

\text{Area}=rec\tan gle_{green}+triangle_{gren}-triangle_{red}\begin{gathered} \text{the area of a rectangle is given by} \\ A_{rec}=lenght\cdot widht \\ \text{and} \\ \text{the area of a triangle is given by:} \\ A_{tr}=\frac{base\cdot height}{2} \end{gathered}

so

\begin{gathered} \text{Area}=rec\tan gle_{green}+triangle_{gren}-triangle_{red} \\ \text{replace} \\ \text{Area}=(F_0\cdot d)+\frac{(F_0\cdot d)}{2}-\frac{(F_0\cdot d)}{2} \\ \text{Area}=(F_0\cdot d) \\ Area=F_0d \end{gathered}

therefore, the answer is

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