If a projectile is shot at a velocity of #9 m/s# and an angle of #pi/6#, how far will the projectile travel before landing?

Question

Lily Adams · Accepted Answer

7.1m

Let the velocity of projection of the object be u with angle of projection #alpha# with the horizontal direction. The vertical component of the velocity of projection is #usinalpha# and the horizontal component is #ucosalpha# Now if the time of flight be T then the object will return to the ground after T sec and during this T sec its total vertical dosplacement h will be zero. So applying the equation of motion under gravity we can write #h=usinalpha xxT+1/2gT^2 =>0=uxxT-1/2xxgxxT^2# where #g="acceleration due to gravity"# #:.T=(2usinalpha)/g# The horizontal displacement during this T sec is #R=ucosalpha xx T=(u^2sin(2alpha))/g# In our problem #u=9m/s;alpha=pi/6 "and" g=9.8m/s^2# Thus, the projectile's flight distance prior to landing is #"Distance"=(9^2sin(2*pi/6))/9.8=7.1m#

Adrian Ayala · Answer

undefined To find the horizontal distance traveled by the projectile, you can use the horizontal component of the initial velocity and the time of flight. The horizontal component of the initial velocity is $ v_{0x} = v_0 \cdot \cos(\theta) $, where $ v_0 $ is the initial velocity and $ \theta $ is the launch angle.

Given:
- Initial velocity, $ v_0 = 9 $ m/s
- Launch angle, $ \theta = \frac{\pi}{6} $

The horizontal component of the initial velocity:
$$ v_{0x} = 9 \cdot \cos\left(\frac{\pi}{6}\right) = 9 \cdot \cos\left(30^\circ\right) = 9 \cdot \frac{\sqrt{3}}{2} = \frac{9\sqrt{3}}{2} \approx 7.794 \text{ m/s} $$

Now, you need to find the time of flight ($ t $) using the vertical motion equation:
$$ h = v_{0y}t - \frac{1}{2}gt^2 $$

Where:
- $ h $ is the maximum height (for maximum height, vertical velocity $ v_{0y} = 0 $)
- $ g $ is the acceleration due to gravity (approximately $ 9.8 $ m/s²)

For maximum height, the equation becomes:
$$ 0 = v_{0y}t - \frac{1}{2}gt^2 $$

Solving for $ t $:
$$ t = \frac{2v_{0y}}{g} $$

The vertical component of the initial velocity:
$$ v_{0y} = v_0 \cdot \sin(\theta) $$
$$ v_{0y} = 9 \cdot \sin\left(\frac{\pi}{6}\right) = 9 \cdot \sin\left(30^\circ\right) = 9 \cdot \frac{1}{2} = 4.5 \text{ m/s} $$

Substitute the values:
$$ t = \frac{2 \times 4.5}{9.8} \approx 0.918 \text{ seconds} $$

Now, you can find the horizontal distance traveled ($ d $) using the formula:
$$ d = v_{0x} \cdot t $$
$$ d = \left(\frac{9\sqrt{3}}{2}\right) \times 0.918 \approx 6.749 \text{ meters} $$

Therefore, the projectile will travel approximately $ 6.749 $ meters before landing.