If a projectile is shot at a velocity of #15 m/s# and an angle of #pi/6#, how far will the projectile travel before landing?

1 Answer
May 16, 2018

The horizontal distance is #=19.88m#

Explanation:

The trajectory of a projectile in the #x-y# plane is given by the equation

#y=xtantheta-(gx^2)/(2u^2cos^2theta)#

The initial velocity is #u=15ms^-1#

The angle is #theta=pi/6#

The acceleration due to gravity is #g=9.8ms^-2#

The vertical distance travelled is #y=0#

Therefore,

#xtan(pi/6)-(9.8x^2)/(2*15^2cos^2(pi/6))=0#

#0.577x-0.029x^2=0#

#x(0.577-0.029x)=0#

#x=0# which is the initial point

#x=0.577/0.029=19.88m#

graph{0.577x-0.029x^2 [-0.96, 21.54, -4.6, 6.65]}