If a projectile is shot at a velocity of #15 m/s# and an angle of #pi/12#, how far will the projectile travel before landing?

Question

Adrian Ayala · Accepted Answer

The distance is #=11.48m#

Resolving in the vertical direction #uarr^+#

initial velocity is #u_y=vsintheta=15*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=15sin(1/12pi)-g*t#

#t=15/(g)*sin(1/12pi)#

#=0.396s#

To find the horizontal distance, we apply the equation of motion

#s=2u_x*t#

#=2*15cos(1/12pi)*0.396#

#=11.48m#

Charlotte Aaron · Answer

undefined To find the horizontal distance traveled by the projectile, you can use the formula:

$$ \text{Range} = \frac{v^2 \cdot \sin(2\theta)}{g} $$

Where:
- $ v $ is the initial velocity of the projectile (15 m/s)
- $ \theta $ is the launch angle (π/12 radians)
- $ g $ is the acceleration due to gravity (9.8 m/s²)

Plugging in the values:

$$ \text{Range} = \frac{(15)^2 \cdot \sin(2 \cdot \frac{\pi}{12})}{9.8} $$

$$ \text{Range} = \frac{225 \cdot \sin(\frac{\pi}{6})}{9.8} $$

$$ \text{Range} = \frac{225 \cdot \frac{1}{2}}{9.8} $$

$$ \text{Range} = \frac{112.5}{9.8} $$

$$ \text{Range} ≈ 11.48 \, \text{m} $$

Therefore, the projectile will travel approximately 11.48 meters before landing.

Adrian Ayala · Answer

undefined To find the horizontal distance traveled by the projectile before landing, we can use the kinematic equations of projectile motion. The horizontal and vertical motions are independent of each other.

The horizontal component of the initial velocity (Vx) is given by:
Vx = V * cos(θ)

Where:
V = initial velocity (15 m/s)
θ = launch angle (π/12)

Substituting the given values:
Vx = 15 * cos(π/12)
Vx ≈ 15 * 0.9659
Vx ≈ 14.4885 m/s

Now, we can use the horizontal component of the initial velocity and the time of flight to find the horizontal distance traveled (range).

The time of flight (T) can be calculated using the vertical component of the initial velocity (Vy) and the acceleration due to gravity (g = 9.8 m/s^2):
Vy = V * sin(θ)

Substituting the given values:
Vy = 15 * sin(π/12)
Vy ≈ 15 * 0.2588
Vy ≈ 3.882 m/s

Now, we can use the equation for time of flight:
T = 2 * Vy / g

Substituting the known values:
T = 2 * 3.882 / 9.8
T ≈ 0.7919 s

Finally, we can find the horizontal distance traveled (range) using the formula:
Range = Vx * T

Substituting the known values:
Range ≈ 14.4885 * 0.7919
Range ≈ 11.47 meters

Therefore, the projectile will travel approximately 11.47 meters horizontally before landing.