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Grade:
        At what angle shuld a body be projected with a velocity 24m/s just to pass over the obstacle 14m high at a distance of 24m.
7 years ago

Answers : (2)

Chetan Mandayam Nayakar
312 Points
							

14=24*tan(theta)-(g*)(242)/(2*242cos2(theta))

7 years ago
lokesh soni
37 Points
							

y=xtanθ-gx2sec2θ /2u2

14=24tanθ-10sec2θ*24*24/2*24*24

tanθ=3.8or 1

θ=45degree or 75degree

6 years ago
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