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.

Question

Chetan Mandayam Nayakar · Answer

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

lokesh soni · Answer

y=xtan&theta;-gx 2 sec 2 &theta; /2u 2 14=24tan&theta;-10sec 2 &theta;*24*24/2*24*24 tan&theta;=3.8or 1 &theta;=45degree or 75degree

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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.

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.

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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.

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.

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