If a projectile is shot at a velocity of 23 m/s and an angle of pi/12, how far will the projectile travel before landing?

1 Answer
Jun 11, 2017

27 "m"

Explanation:

We're asked to find the horizontal range of the launched projectile with a known initial velocity.

To find this distance, we first must find the time t when the object lands, using the equation

Deltay = v_(0y)t - 1/2g t^2

where

  • Deltay is the change in height of the projectile, in "m",

  • v_(0y) is the initial y-velocity, in "m"/"s". To find this, we must use the equation

v_(0y) = v_0sinalpha

where

  • v_0 is the initial speed (23"m"/"s"), and

  • alpha is the initial launch angle (pi/12)

Therefore, the initial y-velocity is

v_(0y) = (23"m"/"s")sin(pi/12) = 5.95"m"/"s"

  • t is the time, in "s", and

  • g is the acceleration due to gravity near Earth's surface, 9.8"m"/("s"^2)

Plugging in 0 for Deltay (because we must find the time when it lands at the same height as it is launched), we have

0 = (5.95"m"/"s")t - 1/2(9.8"m"/("s"^2))t^2

1/2(9.8"m"/("s"^2))t = 5.95"m"/"s"

t = color(red)(1.21 color(red)("s"

Now, to find how far horizontally it traveled, we use the equation

Deltax = v_(0x)t

We must find the initial x-velocity now, using the equation

v_(0x) = v_0cosalpha = 23"m"/"s"cos(pi/12) = 22.2"m"/"s"

The horizontal range is thus

Deltax = (22.2"m"/cancel("s"))(color(red)(1.21)cancel(color(red)("s"))) = color(blue)(27 color(blue)("m"

rounded to 2 significant figures, the amount given in the problem.