Question

A projectile fired from a gun has initial horizontal and vertical components of velocity equal to 60 m/s and 80 m/s, respectively. (a) Determine the initial speed of the projectile. (b) At what angle is the projectile fired (measured with respect to the horizontal)

Answer 1

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°".

(a) LAUNCH SPEED

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,

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

(b) LAUNCH ANGLE

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