A projectile is shot at an angle of #pi/4 # and a velocity of # 2 m/s#. How far away will the projectile land?

1 Answer
Feb 4, 2016

I got #"Distance" ~=0.408" metres"#
To give a full explanation of the calculation method proved too long. So I have given a summery of the method I used.

Explanation:

#color(blue)("Target: Time of flight x horizontal component of projectile")#

Let total time of flight be #T_f#

Find time of flight using

Vertical component of projectile# = 2xxsin(pi/4) -> 2/sqrt(2)#
The vertical travel distance (upwards) at #T_f -> 2/sqrt(2)xxT_f#

This must match the downward distance due to gravity at#" " T_f#.

This is the mean velocity (downwards) multiplied by flight time

#" "(9.81xxT_f)/2 xxT_f#

Equate them to each other and you have a quadratic that can be solved using the standard form equation.

#" I got "T_f~=0.2883# seconds
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Having found #T_f# you then substitute into the distance equation of the horizontal component of the projectile.

#2xxcos(pi/4)xxT_f" " ->" "(2T_f)/sqrt(2) #

#" Distance" ~=0.408" metres"#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Comment")#

#color(brown)("The "underline("actual")" height at any time "T_i" where "0<=T_i<=T_f)# is

#(2/sqrt(2)xxT_i )-((9.81xxT_i^2)/2 )#