A projectile is shot at an angle of #pi/3 # and a velocity of # 6 m/s#. How far away will the projectile land?

Question

A projectile is shot at an angle of #pi/3   # and a velocity of # 6   m/s#. How far away will the projectile land?

Isabella Alvarez · Accepted Answer

The projectile will land 3.2 meters away from its starting point.

I will give an in-depth explanation, as students tend to have some conceptual issues with these types of problems.

You are given the initial velocity and launch angle of a projectile. The angle indicates that you will need to break the velocity up into components, #v_x# and #v_y#. This can be done using basic trigonometry, where the angle of our triangle in question is #pi/3=60^o#.

We know that #v=6m/s#, we want to find #v_x# and #v_y#.

Because we know that #cos(theta)=x/r# (adjacent/hypotenuse), we can find #v_x# by solving for #x#. Thus,

#x=rcos(theta)#

where #r# is equal to #v#, our given initial velocity, and #theta# is equal to #pi/3#.

#v_x=6cos(pi/3)#

#v_x=(3m)/s#

Similarly, we can find #v_y# using #sin(theta)=y/r# (opposite/hypotenuse).

#v_y=rsin(theta)#

#v_y=6sin(pi/3)#

#v_y=(3sqrt(3)m)/s#

Now we have our horizontal and vertical velocities, and we want to solve for horizontal displacement. We can recognize fairly quickly that we don't have enough information about the horizontal motion in order to simply pick a single kinematic equation and plug in our known values, so another step must be required.

A lot of the time in these types of projectile motion problems, that other step involves calculating the time interval over which the projectile motion takes place, i.e. solving for #t#. We can do this using our vertical velocity, since we know that the acceleration in the y-direction for an object undergoing projectile motion must be equal to #-g#, where #g=(9.8m)/s^2#.

We know our acceleration and initial velocity vertically. This allows us to solve for time using either of two kinematic equations:

#y_1=y_0+v_(iy)t+1/2a_yt^2#

or

#v_(fy)=v_(iy)+a_yt#

While both will yield the same answer, it is preferable to use the second option, which will not require use of the quadratic formula.

Thus, we can rearrange the second equation to solve for flight time using algebra. We will use that our final vertical velocity is equal to 0, calculating the projectile's rise time (the projectile momentarily stops when reaching its maximum altitude before falling back down).

#t=(v_(fy))/t#

#t=(-3sqrt(3))/(-9.8)#

#t=0.53s#

Note: I have chosen not to include units in the above calculation as they make it difficult to read, but they are present. Also, note that I have defined the final y velocity as negative because the projectile will be falling toward the earth at the end of its motion, i.e. headed in the negative direction. This yields a favorable net-positive answer for time when we divide by #-g#.

This gives us the rise time of the projectile, i.e. how long it takes to reach its maximum altitude. Assuming the projectile begins and ends at the same height, the rise and fall times are equal. Thus, the total flight time of the projectile is #0.53 * 2 = 1.06s#.

Because time does not have x and y components, this time also applies in the x-direction. The only kinematic option to solve for horizontal displacement given our known values is:

#x_1=x_0+v_(ix)t+1/2a_xt^2#

We must recall that in projectile motion, there is no #a_x#, only #a_y#. Thus, #a_x=0#. Either of the other two common kinematic equations would result in #v_0=v_1#, which, while true (no acceleration), is not very useful to us. Note that, because this is 0 acceleration, we are essentially employing the basic relationship #v=(Δx)/(Δt)#, where you can solve for #x# as #x=vt#.

If you choose to use the kinematic above, because #a_x=0#, the last part of the equation drops away, as does #x_0#, which we can also define as 0.

Thus,

#x_1=v_(ix)t#.

#x=(3m)/s(1.06s)#

#x=3.2m#

Hope that helped!

Carson Alexander · Answer

undefined To determine how far away the projectile will land, we can use the horizontal component of the initial velocity and the time it takes for the projectile to reach the ground. Using the equation:

$$ \text{Horizontal distance} = \text{Horizontal velocity} \times \text{Time of flight} $$

We first find the horizontal component of the initial velocity:

$$ \text{Horizontal velocity} = v_0 \times \cos(\theta) $$

Where $ v_0 $ is the initial velocity and $ \theta $ is the angle of projection. Given $ v_0 = 6 $ m/s and $ \theta = \frac{\pi}{3} $, we have:

$$ \text{Horizontal velocity} = 6 \times \cos\left(\frac{\pi}{3}\right) $$

$$ \text{Horizontal velocity} = 6 \times \frac{1}{2} $$

$$ \text{Horizontal velocity} = 3 \text{ m/s} $$

Next, we need to find the time of flight, which is the time it takes for the projectile to reach the ground. We can use the vertical motion equation:

$$ h = v_{0y} t - \frac{1}{2} g t^2 $$

Where $ h $ is the initial height (assuming it is shot from the ground), $ v_{0y} $ is the vertical component of the initial velocity, $ g $ is the acceleration due to gravity (approximately $ 9.8 \, \text{m/s}^2 $), and $ t $ is the time of flight.

Since the projectile starts from the ground, $ h = 0 $, and $ v_{0y} = v_0 \times \sin(\theta) $:

$$ v_{0y} = 6 \times \sin\left(\frac{\pi}{3}\right) $$

$$ v_{0y} = 6 \times \frac{\sqrt{3}}{2} $$

$$ v_{0y} = 3\sqrt{3} \text{ m/s} $$

Now, we can solve for $ t $:

$$ 0 = (3\sqrt{3}) t - \frac{1}{2} \times 9.8 \times t^2 $$

$$ 0 = 3\sqrt{3} t - 4.9 t^2 $$

Solving this quadratic equation for $ t $, we find $ t \approx 1.58 $ seconds.

Finally, we can calculate the horizontal distance:

$$ \text{Horizontal distance} = 3 \times 1.58 $$

$$ \text{Horizontal distance} \approx 4.74 \text{ meters} $$

So, the projectile will land approximately 4.74 meters away from the launch point.