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Dvinal [7]
3 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]3 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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A model rocket is launched directly upward at a speed of 16 meters per second from a height of 2 meters. The function f(t)=−4.9t
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