A projectile is shot from the ground at a velocity of #7 m/s# at an angle of #pi/12#. How long will it take for the projectile to land?

1 Answer
May 23, 2018

Answer:

The time is #=0.37s#

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=7ms^-1#

The angle is #theta=(1/12pi)rad#

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

The distance #y=0#

Therefore,

#xtan(1/12pi)-(9.8*x^2)/(2*7^2cos^2(1/12pi))=0#

#0.268x-0.107x^2=0#

#x(0.268-0.107x)=0#

#x=0#, this is the starting point

#x=0.268/0.107=2.5m#

The time is #t=x/v_x=2.5/(7cos(1/12pi))=0.37s#