Question

a body is projected from the ground with an initial velocity 10 m/s if g= 10 m/s max range of that projectile will be​

Answer 1

Answer:

Velocity of projectile  v=3i^+10j^   m/s

Speed of projectile u=∣v∣=102+32=109   m/s

Angle of projectile  tanθ=310

⟹sinθ=10910  and  cosθ=1093

Maximum height attained H=2gu2sin2θ

∴H=2(10)109×109100=5m

Horizontal range R=gu2sin2θ=g2u2sinθcosθ

⟹R=102(109)×109