A projectile is shot from the ground at an angle of #pi/6 # and a speed of #5 /9 m/s#. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

Question

A projectile is shot from the ground at an angle of #pi/6  # and a speed of #5  /9     m/s#. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

Charles Aeschliman · Accepted Answer

The distance is #=0.014m#

Resolving in the vertical direction #uarr^+#

initial velocity is #u_y=vsintheta=5/9*sin(1/6pi)#

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=5/9sin(1/6pi)-g*t#

#t=5/(9g)*sin(1/6pi)#

#=0.028s#

Resolving in the horizontal direction #rarr^+#

To find the distance, we apply the equation of motion

#s=u_x*t#

#=5/9cos(1/6pi)*0.028#

#=0.014m#

Isabella Alvarez · Answer

undefined To find the distance from the starting point when the projectile reaches its maximum height, we can use the following steps:

1. Calculate the initial vertical velocity ($V_{i_y}$) using the given angle and speed.
2. Use the kinematic equation for vertical motion to find the time ($t$) it takes for the projectile to reach its maximum height.
3. Use the time obtained in step 2 to find the horizontal distance ($d$) traveled by the projectile using the kinematic equation for horizontal motion.

Given:
- Angle of projection ($\theta$) = $\frac{\pi}{6}$
- Initial speed ($V_i$) = $\frac{5}{9}$ m/s

1. $V_{i_y} = V_i \times \sin(\theta)$
   $V_{i_y} = \frac{5}{9} \times \sin\left(\frac{\pi}{6}\right)$
   $V_{i_y} = \frac{5}{9} \times \frac{1}{2}$
   $V_{i_y} = \frac{5}{18}$ m/s

2. Use the equation:
   $V_f = V_{i_y} + gt$
   At maximum height, $V_f = 0$
   $0 = \frac{5}{18} - 9.8t$
   Solve for $t$:
   $t = \frac{5}{18 \times 9.8}$

3. Use the equation:
   $d = V_{i_x} \times t$
   $V_{i_x} = V_i \times \cos(\theta)$
   $V_{i_x} = \frac{5}{9} \times \cos\left(\frac{\pi}{6}\right)$
   $V_{i_x} = \frac{5}{9} \times \frac{\sqrt{3}}{2}$
   $V_{i_x} = \frac{5\sqrt{3}}{18}$ m/s
   $d = \frac{5\sqrt{3}}{18} \times \frac{5}{18 \times 9.8}$

Calculating $d$:
$d = \frac{5\sqrt{3}}{18} \times \frac{5}{18 \times 9.8}$

Therefore, the distance from the starting point when the projectile reaches its maximum height is approximately 0.063 meters.

Sebastian Acosta · Answer

undefined The projectile's distance from the starting point when it reaches its maximum height is $ \frac{25}{27} $ meters.