A projectile is shot from the ground at an angle of #pi/6 # and a speed of #15 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 #15   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?

Olivia Askew · Accepted Answer

#t=(15*0,5)/(9,81)=0,76" "s#
#x~=9,87 m#
#y=h_m~=2,87 m#
#v_y=0" at the point of D"#
#v_x=12,99 m/s#

#t=v_i*sin alpha/g" 'elapsed time to the maximum height'"#
#t=(15*0,5)/(9,81)=0,76" "s#
#y=h_m=v_i*t*sin alpha-1/2*g*t^2#
#y=h_m=15*0,76*0,5-1/2*9,81*(0,76)^2#
#y=h_m=5,7-2,83#
#y=h_m~=2,87 m#
#x=v_i*t*cos alpha#
#x=15*0,76*0,866#
#x~=9,87 m#
#v_y=0" at the point of D"#
#v_x=v_i*cos alpha#
#v_x=15*0,866#
#v_x=12,99 m/s" The x component of velocity is not change"#

Charles Aeschliman · Answer

undefined To find the projectile's distance from the starting point at maximum height, use the formula: $$ \text{Horizontal distance} = \frac{v_0^2 \sin(2\theta)}{g} $$
where $ v_0 $ is the initial speed, $ \theta $ is the launch angle, and $ g $ is the acceleration due to gravity. Plugging in the values: $ v_0 = 15 \, \text{m/s} $, $ \theta = \pi/6 $, and $ g = 9.8 \, \text{m/s}^2 $, calculate the horizontal distance.