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vladimir1956 [14]

3 years ago

7

A projectile is launched into the air with an initial speed of 40 m/s and a launch angle of 20° above the horizontal. The projec

tile lands on the ground five seconds later. Neglecting air resistance, calculate the projectile’s range and draw a projectile path.

1 answer:

miv72 [106K]3 years ago

7 0

Explanation:

V=40m/s

Vy=V.sina=40.sin20=40 . 0.342=13.68m/s

Vx=V.cosa=40.cos20=40 . 0.766=30.64m/s

Projectile travels during 5 seconds and the ramge becomes:

x=V.t=30.64 . 5=153.2m

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Dima020 [189]

Explanation:

Given that,

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$u'=\dfrac{0.8c-0.6c}{1-\dfrac{0.6c\times0.8c}{c^2}}$

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Using formula of kinetic energy

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Where, Proton mass energy = m₀c²

Put the value into the formula

$K.E=938.28\times(\dfrac{1}{\sqrt{1-\dfrac{(0.6c)^2}{c^2}}}-1)$

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$K.E=234.57\ MeV$

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Using formula of kinetic energy

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$K.E=938.28\times(\dfrac{1}{\sqrt{1-(0.385)^2}}-1)$

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Using formula of momentum

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Proton mass energy = m₀c²

$m_{0}=\dfrac{938.28}{c^2}$

$P_{obs}=\dfrac{\dfrac{938.28}{c^2}\times(0.385c)^2}{\sqrt{1-\dfrac{(0.385c)^2}{c^2}}}$

$P_{obs}=\dfrac{938.28\times(0.385)^2}{\sqrt{1-0.385^2}}$

$P_{obs}=150.69\ MeV/c$

Hence, This is required solution.

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