# 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?

##### 1 Answer

#### Explanation:

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

**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

# { (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

# :. 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,

So we can calculate

# s=2/5cos(pi/4)T #

# :. s=1/5sqrt(2)T #

# :. s=1/5sqrt(2)sqrt(2)/(5g) #

# :. s=2/(25g) #

So using

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