A projectile is shot from the ground at an angle of #pi/4 # and a speed of #2/5 ms^-1#. 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 #2/5   ms^-1#. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

Charles Aeschliman · Accepted Answer

#8.1mm#

For Physics or Mechanics you should learn the "suvat" equations for motion under constant acceleration:

#{: (v=u+at, " where ", s="displacement "(m)), (s=ut+1/2at^2, , u="initial speed "(ms^-1)), (s=1/2(u+v)t, , v="final speed "(ms^-1)), (v^2=u^2+2as, , a="acceleration "(ms^-2)), (s=vt-1/2at^2, , t="time "(s)) :} #

Vertical Motion
Motion under constant acceleration due to gravity, applied vertically upwards

Let the total time that the projectile takes to its maximum point be #T#, at the point the velocity will momentary be zero.

# { (s=,"not required",m),(u=,2/5 sin(pi/4)=1/5sqrt(2),ms^-1),(v=,0,ms^-1),(a=,-g,ms^-2),(t=,T,s) :} #

So we can calculate #T# using #v=u+at#

# :. 0=1/5sqrt(2)-gT #
# :. gT=1/5sqrt(2) #
# :. T=sqrt(2)/(5g)#

Horizontal Motion
Under constant speed (NB we can still use "suvat" equation with a=0). Thje projectile will be in in the air for the same time, #t =T#

So we can calculate #s# using #s=ut#

# s=2/5cos(pi/4)T #
# :. s=1/5sqrt(2)T #
# :. s=1/5sqrt(2)sqrt(2)/(5g) #
# :. s=2/(25g) #

So using #g=9.8 ms^-2# we have.

#s=2/245 = 0.008163...= 0.0081 m#, or #8.1mm#

Nathan Ashcroft · Answer

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

$$ d = v_{0x} \cdot t $$

where $ v_{0x} $ is the initial velocity in the horizontal direction and $ t $ is the time it takes for the projectile to reach its maximum height. The horizontal component of the initial velocity can be calculated as $ v_{0x} = v_0 \cdot \cos(\theta) $, where $ v_0 $ is the initial speed and $ \theta $ is the angle of projection.

Given:
- Initial speed ($ v_0 $) = $ \frac{2}{5} $ m/s
- Angle of projection ($ \theta $) = $ \frac{\pi}{4} $

Calculate $ v_{0x} $ using $ v_{0x} = v_0 \cdot \cos(\theta) $:
$$ v_{0x} = \frac{2}{5} \cdot \cos\left(\frac{\pi}{4}\right) $$
$$ v_{0x} = \frac{2}{5} \cdot \frac{\sqrt{2}}{2} $$
$$ v_{0x} = \frac{\sqrt{2}}{5} $$

The time taken to reach maximum height ($ t $) can be calculated using the formula for vertical motion:
$$ t = \frac{v_{0y}}{g} $$

where $ v_{0y} $ is the initial velocity in the vertical direction and $ g $ is the acceleration due to gravity (approximately $ 9.8 \, \text{m/s}^2 $). Since the projectile reaches its maximum height, its vertical velocity ($ v_{0y} $) at that point is 0.

Calculate $ t $ using $ t = \frac{v_{0y}}{g} $:
$$ v_{0y} = v_0 \cdot \sin(\theta) $$
$$ v_{0y} = \frac{2}{5} \cdot \sin\left(\frac{\pi}{4}\right) $$
$$ v_{0y} = \frac{2}{5} \cdot \frac{\sqrt{2}}{2} $$
$$ v_{0y} = \frac{\sqrt{2}}{5} $$

$$ t = \frac{\frac{\sqrt{2}}{5}}{9.8} $$
$$ t = \frac{\sqrt{2}}{49} $$

Now, use $ d = v_{0x} \cdot t $:
$$ d = \frac{\sqrt{2}}{5} \cdot \frac{\sqrt{2}}{49} $$
$$ d = \frac{2}{49} $$

So, the projectile's distance from the starting point when it reaches its maximum height is $ \frac{2}{49} $ meters.