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Fantom [35]
2 years ago
8

A projectile is launched at an angle of 45 degrees with an initial speed of 32 feet per second. How far does the projectile trav

el?
Mathematics
1 answer:
Hatshy [7]2 years ago
6 0

Answer:

As the first part comes to the rest means, then it will fall under gravity directly.

First of all maximum height,

H=

g

u²sin²θ

H=

2×10

400×(

2

1

)

H=10m.

As initially the full projectile was supposed to land at A but due to explosion first part land at A', so to make the centre of mass lies at A the second part should lie at B.

f they don't break (explode) range OA

OA=

g

u²sin2θ

OA=400×sin45×

10

2

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Answer:

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Step-by-step explanation:

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1) 0.01

y(1 + 0.01)=y(1.01) = 43(1.01)-23(1.01)^{2} = 19.9677\\\\\text{Average Velocity} = \dfrac{y(1.01)-y(1)}{1.01-1} =\dfrac{19.9677-20}{1.01-1}= \dfrac{-0.0323}{0.01} = -3.230000 \text{feet per second}

2) 0.005 s

y(1 + 0.005)=y(1.005) = 43(1.005)-23(1.005)^{2} = 19.984425\\\\\text{Average Velocity} = \dfrac{y(1.005)-y(1)}{1.005-1} =\dfrac{19.984425-20}{1.005-1} = -3.1150000 \text{feet per second}

3) 0.002 s

y(1 + 0.002)=y(1.002) = 43(1.002)-23(1.002)^{2} = 19.993908\\\\\text{Average Velocity} = \dfrac{y(1.002)-y(1)}{1.002-1} =\dfrac{19.993908-20}{1.002-1} = -3.0460000 \text{feet per second}

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