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vichka [17]
3 years ago
5

A tennis player standing 12.6 m from the net hits the ball at 3.18° above the horizontal. To clear the net, the ball must rise a

t least 0.362 m. If the ball just clears the net at the apex of its trajectory, how fast was the ball moving when it left the racket?
Physics
1 answer:
SOVA2 [1]3 years ago
3 0
At the vertex, it's vertical velocity is 0, since it has stopped moving up and is about to come back down, and its displacement is 0.33m.
So we use v² = u² + 2as (neat trick I discovered just then for typing the squared sign: hold down alt and type 0178 on ur numpad wtih numlock on!!!) ANYWAY.......
We apply v² = u² + 2as in the y direction only. Ignore x direction.
IN Y DIRECTION:
v² = u² + 2as
0 = u² - 2gh
u = √(2gh) (Sub in values at the very end)
So that will be the velocity in the y direction only. But we're given the angle at which the ball is hit (3° to the horizontal). So to find the velocity (sum of the velocity in x and y direction on impact) we can use: sin 3° = opposite/hypotenuse = (velocity in y direction only) / (velocity)
So rearranging,
velocity = (velocity in y direction only) / sin 3°
= √(2gh)/sin 3°
= (√(2 x 9.8 x 0.33)) / sin 3°
= 49 m/s at 3° to the horizontal
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A Tennis ball falls from a height 40m above the ground the ball rebounds
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If the ball is dropped with no initial velocity, then its velocity <em>v</em> at time <em>t</em> before it hits the ground is

<em>v</em> = -<em>g t</em>

where <em>g</em> = 9.80 m/s² is the magnitude of acceleration due to gravity.

Its height <em>y</em> is

<em>y</em> = 40 m - 1/2 <em>g</em> <em>t</em>²

The ball is dropped from a 40 m height, so that it takes

0 = 40 m - 1/2 <em>g</em> <em>t</em>²

==>  <em>t</em> = √(80/<em>g</em>) s ≈ 2.86 s

for it to reach the ground, after which time it attains a velocity of

<em>v</em> = -<em>g</em> (√(80/<em>g</em>) s)

==>  <em>v</em> = -√(80<em>g</em>) m/s ≈ -28.0 m/s

During the next bounce, the ball's speed is halved, so its height is given by

<em>y</em> = (14 m/s) <em>t</em> - 1/2 <em>g</em> <em>t</em>²

Solve <em>y</em> = 0 for <em>t</em> to see how long it's airborne during this bounce:

0 = (14 m/s) <em>t</em> - 1/2 <em>g</em> <em>t</em>²

0 = <em>t</em> (14 m/s - 1/2 <em>g</em> <em>t</em>)

==>  <em>t</em> = 28/<em>g</em> s ≈ 2.86 s

So the ball completes 2 bounces within approximately 5.72 s, which means that after 5 s the ball has a height of

<em>y</em> = (14 m/s) (5 s - 2.86 s) - 1/2 <em>g</em> (5 s - 2.86 s)²

==>  (i) <em>y</em> ≈ 7.5 m

(ii) The ball will technically keep bouncing forever, since the speed of the ball is only getting halved each time it bounces. But <em>y</em> will converge to 0 as <em>t</em> gets arbitrarily larger. We can't realistically answer this question without being given some threshold for deciding when the ball is perfectly still.

During the first bounce, the ball starts with velocity 14 m/s, so the second bounce begins with 7 m/s, and the third with 3.5 m/s. The ball's height during this bounce is

<em>y</em> = (3.5 m/s) <em>t</em> - 1/2 <em>g</em> <em>t</em>²

Solve <em>y</em> = 0 for <em>t</em> :

0 = (3.5 m/s) <em>t</em> - 1/2 <em>g t</em>²

0 = <em>t</em> (3.5 m/s - 1/2 <em>g</em> <em>t</em>)

==>  (iii) <em>t</em> = 7/<em>g</em> m/s ≈ 0.714 s

As we showed earlier, the ball is in the air for 2.86 s before hitting the ground for the first time, then in the air for another 2.86 s (total 5.72 s) before bouncing a second time. At the point, the ball starts with an initial velocity of 7 m/s, so its velocity at time <em>t</em> after 5.72 s (but before reaching the ground again) would be

<em>v</em> = 7 m/s - <em>g t</em>

At 6 s, the ball has velocity

(iv) <em>v</em> = 7 m/s - <em>g</em> (6 s - 5.72 s) ≈ 4.26 m/s

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(a) Power is given by :

P=V\times I\\\\I=\dfrac{P}{V}\\\\I=\dfrac{1650\ W}{110\ V}\\\\I=15\ A

(b) Again the power is given by :

P=\dfrac{V^2}{R}

R is resistance of the dryer

R=\dfrac{V^2}{P}\\\\R=\dfrac{(110)^2}{1650}\\\\R=7.34\ \Omega

(c) The magnetic field produced by the dryer at the user's hand is given by :

B=\dfrac{\mu_o I}{2\pi r}\\\\B=\dfrac{4\pi \times 10^{-7}\times 15}{2\pi \times 1.95\times 10^{-2}}\\\\B=1.53\times 10^{-4}\ T

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3 years ago
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