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Lera25 [3.4K]
3 years ago
11

A second baseman tosses the ball to the first baseman, who catches it at the same level from which it was thrown. The throw is m

ade with an initial speed of 18.0m/s at an angle of 37.5 above the horizontal. A) What is the horizontal component of the ball's velocity just before it is caught? B) How long is the ball in the air?. . Can someone please explain to me how this works not just the answer I'm somewhat confused.
Physics
1 answer:
Anit [1.1K]3 years ago
6 0
 <span>(a) 

Taking the angle of the pitch, 37.5°, and the particle's initial velocity, 18.0 ms^-1, we get: 

18.0*cos37.5 = v_x = 14.28 ms^-1, the projectile's horizontal component. 

(b) 
To much the same end do we derive the vertical component: 

18.0*sin37.5 = v_y = 10.96 ms^-1 

Which we then divide by acceleration, a_y, to derive the time till maximal displacement, 

10.96/9.8 = 1.12 s 

Finally, doubling this value should yield the particle's total time with r_y > 0 

<span>2.24 s

I hope my answer has come to your help. Thank you for posting your question here in Brainly.
</span></span>
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Answer:

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Given that,

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(A). We need to calculate the energy of the photon

Using formula of rest mass energy

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E=9.109\times10^{-31}\times(3\times10^{8})^2

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E=512375\ eV

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The total energy of photon

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Using formula of wavelength

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\lambda=\dfrac{6.626\times10^{−34}\times3\times10^{8}}{1.546\times10^{6}\times1.6\times10^{-19}}

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