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

Julia Aguilar · Accepted Answer

The distance is #=0.92m#

Resolving in the vertical direction #uarr^+#

initial velocity is #u_y=vsintheta=4*sin(1/4pi)#

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=4sin(1/4pi)-g*t#

#t=4/g*sin(1/4pi)#

#=0.29s#

The greatest height is

#h=(4sin(pi/4))^2/(2g)=0.41m#

Resolving in the horizontal direction #rarr^+#

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

#s=u_x*t#

#=4cos(1/4pi)*0.29#

#=0.82m#

The distance from the starting point is

#d=sqrt(h^2+s^2)#

#=sqrt(0.41^2+0.82^2)#

#=0.92m#

Charlotte Aaron · Answer

undefined When a projectile reaches its maximum height, its vertical velocity component becomes zero. The time taken to reach maximum height can be calculated using the vertical component of the initial velocity and the acceleration due to gravity. The horizontal distance traveled during this time can be calculated using the horizontal component of the initial velocity and the time taken. Using the equation $d = v_i t + \frac{1}{2} a t^2$ for vertical motion and $d = v_i t$ for horizontal motion, where $d$ is distance, $v_i$ is initial velocity, $a$ is acceleration, and $t$ is time, you can find the horizontal distance from the starting point when the projectile reaches its maximum height.