A projectile is shot from the ground at a velocity of #3 m/s# at an angle of #pi/3#. How long will it take for the projectile to land?

1 Answer
Apr 25, 2018

Answer:

The time is #=1.06s#

Explanation:

The equation describing the trajectory of the projectile in the #x-y# plane is

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

The initial velocity is #u=3ms^-1#

The angle is #theta=(1/3pi)rad#

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

The distance #y=0#

Therefore,

#xtan(1/3pi)-(9.8*x^2)/(2*3^2cos^2(pi/3))=0#

#1.732x-1.089x^2=0#

#x(1.732-1.089x)=0#

#x=0#, this is the starting point

#x=1.732/1.089=1.59m#

The time is #t=x/v_x=1.59/(3cos(pi/3))=1.06s#