A projectile is shot from the ground at an angle of #(5 pi)/12 # and a speed of #3/5 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 #(5 pi)/12 # and a speed of #3/5    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?

Emily Abbate · Accepted Answer

#"distance" = 0.0309# #"m"#

Given a particle's initial velocity, we are asked to determine how far it is from the launch point at the moment of its maximum height.

We can use the equations to accomplish this.

#(v_y)^2 = (v_0sinalpha_0)^2 - 2g(Deltay)#

and

#Deltax = v_0cosalpha_0t#

to determine the particle's maximum height in both vertical and horizontal directions.

We realize that the instantaneous s#y#-velocity #v_y# is #0# at its maximum height, so we have

#0 = (3/5color(white)(l)"m/s")sin((5pi)/12) - 2(9.81color(white)(l)"m/s"^2)(Deltay)#

#Deltay = color(red)(0.0295# #color(red)("m"#

The duration required for this to occur is provided by

#v_y = v_0sinalpha_0 - g t#

#0 = (3/5color(white)(l)"m/s")sin((5pi)/12) - (9.81color(white)(l)"m/s"^2)t#

#t = 0.0591# #"s"#

We can now use the second equation to determine the horizontal distance #Deltax#:

#Deltax = v_0cosalpha_0t = (3/5color(white)(l)"m/s")cos((5pi)/12)(0.0591color(white)(l)"s")#

#Deltax = color(green)(0.00917# #color(green)("m"#

Consequently, the separation from the launch site is

#r = sqrt((Deltax)^2 + (Deltay)^2) = sqrt((color(green)(0.00917color(white)(l)"m"))^2 + (color(red)(0.0295color(white)(l)"m"))^2)#

#= color(blue)(ul(0.0309color(white)(l)"m"#

Nicholas Agrusa · Answer

undefined To find the distance from the starting point when the projectile reaches its maximum height, you can use the formula:

$$ \text{distance} = \text{initial velocity} \times \text{time at maximum height} \times \cos(\text{launch angle}) $$

First, find the time it takes for the projectile to reach its maximum height:

$$ \text{time to maximum height} = \frac{\text{initial vertical velocity}}{\text{vertical acceleration}} $$

Then, use this time to calculate the horizontal distance:

$$ \text{distance} = \text{initial horizontal velocity} \times \text{time to maximum height} \times \cos(\text{launch angle}) $$

$$ \text{distance} = \left( \frac{3}{5} \right) \times \left( \frac{\frac{3}{5} \sin\left(\frac{5\pi}{12}\right)}{-9.8} \right) \times \cos\left(\frac{5\pi}{12}\right) $$

$$ \text{distance} ≈ 0.166 \, \text{m} $$