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MariettaO [177]

3 years ago

14

A football player punts the ball from the ground at a 65.0° angle above the horizontal. If the ball stays in the air for a total

of 6.5 seconds, what are the vertical and horizontal components of the initial velocity?

1 answer:

maw [93]3 years ago

5 0

Answer: $V_{0,x } = 297.7 \frac{m}{s}\\V_{0,y } = 637 \frac{m}{s}$

Explanation:

Hi!

We define the point (0,0) as the intial position of the ball. The initial velocity is $(V_{0,x}, V_{0,y})$

The motion of the ball in the horizontal direction (x) has constant velocity, because there is no force in that direction. :

$x(t) = V_{0,x}t$

In the vertical direction (y), there is the downward acceleration g of gravity:

$y(t) = -gt^2 + V_{0,y}t$

(note the minus sign of acceleration, because it points in the negative y-direction)

When the ball hits the ground, at t = 65s, y(t = 65 s) = 0. We use this to find the value of the initial vertical velocity:

$0 = t(-gt + V_{0,y})\\V_{0,y} = gt = 9.8 \frac{m}{s^2} 65 s = 637 \frac{m}{s}$

We used that g = 9.8 m/s²

To find the horizonttal component we use the angle:

$\tan(65\º) = \frac{V_{y,0}}{V_{x,0}} = 2.14\\V_{x,0} = 297.7\frac{m}{s}$

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Explanation:

Given:

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Final speed of the stone is 0 m/s as it comes to rest.

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Total time taken is, $t_{total}=17.5\ s$

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Now, we know that, average speed can also be expressed as:

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$\mu_dmg=ma\\\mu_d=\frac{a}{g}$

Plug in 0.244 for 'a' and 9.8 for 'g'. This gives,

$\mu_d=\frac{a}{g}=\frac{0.244}{9.8}=0.025$

Therefore, the coefficient of dynamic friction is 0.025.

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