A projectile is shot at an angle of #pi/12 # and a velocity of # 13 m/s#. How far away will the projectile land?

Question

A projectile is shot at an angle of #pi/12   # and a velocity of # 13    m/s#. How far away will the projectile land?

Emily Abbate · Accepted Answer

The distance is #=8.62m#

Resolving in the vertical direction #uarr^+#

initial velocity is #u_y=vsintheta=13*sin(1/12pi)#

Acceleration is #a=-g#

At the maximum height, #v=0#

We apply the equation of motion

#v=u+at#

to calculate the time to reach the greatest height

#0=13sin(1/12pi)-g*t#

#t=13/g*sin(1/12pi)#

#=0.34s#

Resolving in the horizontal direction #rarr^+#

To find the distance where the projectile will land, we apply the equation of motion

#s=u_x*2t#

#=13cos(1/4pi)*0.34*2#

#=8.62m#

Nora Arzate · Answer

undefined To find the horizontal range of the projectile, you can use the horizontal component of its initial velocity and the time it takes to reach the ground. The horizontal component of velocity can be found using trigonometric functions. The formula for horizontal range is:

Range = (initial velocity * time) * cosine(angle)

In this case, the angle is π/12 and the initial velocity is 13 m/s. Since there is no information about the time, you'll need to calculate it using the vertical component of the velocity and the acceleration due to gravity. The formula for time of flight is:

time = (2 * vertical velocity) / gravity

Then, you can plug the calculated time into the range formula to find the horizontal range. Assuming the projectile is launched from the ground and ignoring air resistance:

time = (2 * initial vertical velocity) / gravity
time = (2 * initial velocity * sine(angle)) / gravity

Substitute the given values:
time = (2 * 13 * sine(π/12)) / 9.8 m/s^2
time ≈ 1.19 seconds

Now, calculate the horizontal range:
Range = (initial velocity * time) * cosine(angle)

Substitute the given values:
Range = (13 * 1.19) * cosine(π/12)
Range ≈ 13.9 meters

So, the projectile will land approximately 13.9 meters away.