Question

a projectile is projected with a velocity 30i+40j m/s its range

Answer 1

  



The velocity of projection of an oblique projectile is (6i+8j)m/s(6i+8j)m/s. The horizontal range of projectile is (g=10m/s)2(g=10m/s)2

(A)4.9m(B)9.6m(C)19.6m(D)14m(A)4.9m(B)9.6m(C)19.6m(D)14m



A)


Answer :(B)9.6m(B)9.6m

v→=6i^+8j^v→=6i^+8j^



Comparing with

v→=vxi^=vyj^v→=vxi^=vyj^ we get,

vx=6ms−1vx=6ms−1

vy=8ms−1vy=8ms−1

Also, v2=v2x+v2yv2=vx2+vy2

=36+64=100=36+64=100

or v=10ms−1v=10ms−1

sinθ=810sin⁡θ=810 and cosθ=610cos⁡θ=610

R=v2sin2θg=2v2sinθcosθgR=v2sin⁡2θg=2v2sin⁡θcos⁡θg

R=2×10×10×810×619×110R=2×10×10×810×619×110mm

=9.6m

Answer 2

answer is range = 240 m

Explanation:

range =u*2sin2 theta%g