Question

When you DO consider air resistance, does the angle for which the range is a maximum depend upon the intial velocity of the projectile? Could you please prove it also? Thanks a lot.

Answer 1

assumption:

air resistance is constant through out the motion.

proof:

consider initial speed to be v and angle to be theta.

let deceleration due to air resistance in vertical direction is a1 and in horizontal direction is a2.

then time taken to reach zero height(i.e. land back on ground) let be t.

using the formula:

displacement=initial speed*time+0.5*acceleration*time^2

==>0=v*sin(theta)*t-0.5*(g+a1)*t^2

==>t=2*v*sin(theta)/(g+a1)

range=horizontal distance travelled

=v*cos(theta)*t-0.5*a2*t^2

=(v^2*sin(2*theta)/(g+a1))-0.5*a2*(4*v^2*sin^2(theta)/(g+a1)^2)

taking derivative w.r.t. theta,

v^2*2*cos(2*theta)/(g+a1)=0.5*a2*4*v^2*sin(2*theta)/(g+a1)^2

==>tan(2*theta)=4*(g+a1)/a2

so depending upon a1 and a2 i.e. acceleration component of air resistance, theta value will change.