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Tems11 [23]
3 years ago
14

A projectile fired from a gun has initial horizontal and vertical components of velocity equal to 60 m/s and 80 m/s, respectivel

y. (a) Determine the initial speed of the projectile. (b) At what angle is the projectile fired (measured with respect to the horizontal)
Physics
1 answer:
DerKrebs [107]3 years ago
6 0

This question involves the concepts of projectile motion and launch speed.

(a) The initial launch speed of the projectile is "100 m/s".

(b) The launch angle of the projectile is "53.13°".

<h3>(a) LAUNCH SPEED</h3>

A projectile motion is a motion that takes place on both x and y axes, simultaneously. In this motion the initial launch speed is given by the following formula:

v_o=\sqrt{v_{ox}^2+v_{oy}^2}

where,

  • v_o = initial launch speed = ?
  • v_{ox} = horizontal component of initial launch speed = 60 m/s
  • v_{oy} = vertical component of initial launch speed = 80 m/s

Therefore,

v_o = \sqrt{(60\ m/s)^2+(80\ m/s)^2}\\\\v_o = 100 m/s

<h3>(b) LAUNCH ANGLE</h3>

Launch angle is given by th following formula:

\theta = tan^{-1}(\frac{v_{oy}}{v_{ox}})=tan^{-1}(\frac{80\ m/s}{60\ m/s})\\\\\theta=53.13^o

Learn more about the projectile motion here:

brainly.com/question/11049671

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