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Dvinal [7]
4 years ago
5

Projectile's horizontal range on level ground is R=v20sin2θ/g. At what launch angle or angles will the projectile land at half o

f its maximum possible range. Enter your answers numerically separated by commas. Express your answer using two significant figures.
Physics
1 answer:
seraphim [82]4 years ago
7 0

Answer:

\theta = 15^o \: or\: 75^o

Explanation:

As we know that the formula of range is given as

R = \frac{v^2sin2\theta}{g}

now we know that

maximum value of the range of the projectile is given as

R_{max} = \frac{v^2}{g}

now we need to find such angles for which the range is half the maximum value

so we will have

\frac{R}{2} = \frac{v^2}{2g} = \frac{v^2sin(2\theta)}{g}

sin(2\theta) = \frac{1}{2}

2\theta = 30 or 150

\theta = 15^o \: or\: 75^o

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What is the acceleration of a wagon of mass 20kg if a horizontal force of 64 is applied to it
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Here we have to calculate the acceleration of the wagon.

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the horizontal force that is applied on the body is 64 newton.

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as per Newton's second law of motion we know that the rate of change of momentum is the applied force.from the quantitative definition of Newton's second law we know that applied force is equal to the product of mass and acceleration.

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let F= 64 N,then a=\frac{64N}{20kg}

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EleoNora [17]

Answer:

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1st part

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