A projectile is shot from the ground at an angle of #( pi)/3 # and a speed of #1 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)/3 # and a speed of #1     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?

Elena Albarran · Accepted Answer

The distance is #=0.044m#

Resolving in the vertical direction #uarr^+#

initial velocity is #u_y=vsintheta=1*sin(1/3pi)#

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=1sin(1/3pi)-g*t#

#t=1*1/g*sin(1/3pi)#

#=0.088s#

Resolving in the horizontal direction #rarr^+#

We apply the equation of motion

#s=u_x*t#

#=1cos(1/3pi)*0.088#

#=0.044m#

Adam Arnold · Answer

undefined The horizontal distance traveled by the projectile when it reaches its maximum height is equal to the horizontal component of its initial velocity multiplied by the time it takes to reach maximum height. Since the horizontal velocity is constant throughout the projectile's motion, the time to reach maximum height is half of the total time of flight.

Given:
Initial speed ($v_0$) = 1 m/s
Launch angle ($\theta$) = $ \frac{\pi}{3} $

Using the horizontal component of the initial velocity:
$$ v_{0x} = v_0 \cos(\theta) $$

To find the time to reach maximum height:
$$ t_{\text{max}} = \frac{v_{0y}}{g} $$

where $ v_{0y} = v_0 \sin(\theta) $ and $ g $ is the acceleration due to gravity (approximately 9.81 m/s²).

The horizontal distance traveled at maximum height is:
$$ d = v_{0x} \cdot t_{\text{max}} $$

Substitute the given values into the equations to find the horizontal distance.