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A man throws a ball at an angle of 45^@ with thehorizontal plane from a height of 15 m. If the shot strikes the ground at a horizontal distance of 30 in, the velocity of throw is (Takeg=10 m s^(-2)) |
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Answer» 10 m `s^(-1)` so, `30=u cos 45^@ t = u/sqrt2t` …(i) For vertical DOWNWARD motion , -y=u sin `45^@ t -1/2 g t^2` or `-15=u/sqrt2 t- 1/2 xx 10 xx t^2` or t=3s (Using (i)) `therefore` Velocity of the THROW , `u=(30sqrt2)/t=10sqrt2 m s^(-1)` |
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