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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.
14=24*tan(theta)-(g*)(242)/(2*242cos2(theta))
y=xtanθ-gx2sec2θ /2u2 14=24tanθ-10sec2θ*24*24/2*24*24 tanθ=3.8or 1 θ=45degree or 75degree
y=xtanθ-gx2sec2θ /2u2
14=24tanθ-10sec2θ*24*24/2*24*24
tanθ=3.8or 1
θ=45degree or 75degree
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