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A projectile is shot at an angle of #pi/8 # and a velocity of # 8 m/s#. How far away will the projectile land?

Answer 1

William Audy

Range = #8.96 x 10 ^ -4# m

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Answer 2

Lily Adams

#"the x-range of projectile is 4.61 m"#

#"the x-range of the projectile can be calculated using formula below."#

#x_m=(v_i^2*sin(2alpha))/g#

#"where ; "#
#v_i=8 m/s#
#g=9.81" " m.s^(-1)#

#alpha=pi/8#

#2*alpha=pi/4" , "sin(pi/4)=0.707#

#x_m=(8^2*0.707)/(9.81)=(64*0.707)/(9.81)#

#x_m=4.61 m#

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Answer 3

Axel Acevedo

To calculate the horizontal range of a projectile, you can use the formula:

Range = (Initial Velocity^2 * sin(2*Theta)) / g

Where:

Initial Velocity is the initial speed of the projectile.
Theta (θ) is the angle at which the projectile is launched.
g is the acceleration due to gravity (approximately 9.81 m/s^2).

In this case:

Initial Velocity (v) = 8 m/s
Angle (θ) = π/8 radians
Acceleration due to gravity (g) = 9.81 m/s^2

Plugging these values into the formula:

Range = (8^2 * sin(2*(π/8))) / 9.81

Calculating the value:

Range = (64 * sin(π/4)) / 9.81

Range = (64 * √2/2) / 9.81

Range = (64 * 0.7071) / 9.81

Range = 45.2564 / 9.81

Range ≈ 4.62 meters

So, the projectile will land approximately 4.62 meters away.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.