A projectile is shot from the ground at an angle of #pi/6 # and a speed of #3/4 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 #3/4 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?

William Audy · Accepted Answer

The distance is #=0.025m#

Solving in the vertical direction #uarr^+#

#u=u_0sintheta=3/4sin(1/6pi)#

#a=-g#

#v=0# at the maximum height

We utilize the formula.

#v=u+at#

#t=(v-u)/a=(0-3/4sin(1/6pi))/(-g)#

#=3/(4g)sin(1/6pi)=0.038s#

Time to reach the maximum height is #=0.038s#

Solving in the horizontal direction #->+#

distance #d=u_0tcos(1/8pi)=0.038*3/4*cos(1/6pi)#

#=0.025m#

Isabella Alvarez · Answer

undefined To find the distance from the starting point when the projectile reaches its maximum height, you can use the following equation: $ d = v_0^2 \cdot \sin(2\theta) / g $, where $ v_0 $ is the initial velocity, $ \theta $ is the launch angle, and $ g $ is the acceleration due to gravity. Plugging in the given values, the initial velocity ($ v_0 = 3/4 $ m/s), launch angle ($ \theta = \pi/6 $), and the acceleration due to gravity ($ g = 9.8 $ m/s²), we get:

$ d = (3/4)^2 \cdot \sin(2 \cdot \pi/6) / 9.8 $

$ d \approx 0.168 $ meters.