Modern Physics> projectile motion(equation of trajectory)...

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.

gaurav shetty , 12 Years ago

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2 Answers

Chetan Mandayam Nayakar

Last Activity: 12 Years ago

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

lokesh soni

Last Activity: 12 Years ago

y=xtanθ-gx²sec²θ/2u²

14=24tanθ-10sec²θ*24*24/2*24*24

tanθ=3.8or 1

θ=45degree or 75degree

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Modern Physics> projectile motion(equation of trajectory)...

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.

gaurav shetty , 12 Years ago

Chetan Mandayam Nayakar

lokesh soni

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