A projectile is shot at an angle of #pi/4 # and a velocity of # 21 m/s#. How far away will the projectile land?

Answer 1

The projectile will land 45 m downfield, assuming it was launched over level ground.

Rather start from scratch (meaning the equations of motion) to solve this one, I will use an equation that can be derived for the range #R# of a projectile:

#R=v_o^2/gsin 2theta#

where #v_o# is the launch speed and #theta# is the launch angle.

It is now just a matter of replacing the variables with values:

#R=21^2/9.8 sin (pi/2) = 45 m#

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Answer 2

Adam Arnold

To find the horizontal distance traveled by the projectile, you can use the formula: range = (initial velocity^2 * sin(2theta)) / g, where theta is the angle of projection and g is the acceleration due to gravity (approximately 9.8 m/s^2). Plugging in the given values: range = (21^2 * sin(2(pi/4))) / 9.8. Simplifying: range ≈ (441 * sin(pi/2)) / 9.8 ≈ (441 * 1) / 9.8 ≈ 44.91 meters. So, the projectile will land approximately 44.91 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.