### Discussion :: Applied Mechanics - Section 2 (Q.No.5)

Abhay said: (Jan 25, 2016) | |

The velocity of a projectile a height h is given by: v = sqrt(u^2 - 2gh). v = sqrt(100.3^2 - 2*9.81*100). v = 90 m/sec. |

Utsav said: (Jul 21, 2016) | |

It is wrong. The correct one is: v = sqrt((u * sin60)^2 - 2gh). v = 75. |

Sachin Singh said: (Dec 31, 2016) | |

How is this right? V = 75. |

Ravi Mehta said: (Jan 8, 2017) | |

As per me, option B is correct. |

Deep said: (Dec 4, 2018) | |

Shouldn't it be like v= √(v * 2ghcotθ)? As the horizontal distance for 60° and 100 m vertical is 100cot60. |

Suresh Reddy said: (Mar 22, 2019) | |

As per when the moves upward it lose their velocity at the rate of(9.8m/S)/S. In vertical direction velocity ---u^2sin^2(60). (V@vertical)^2 = u2sin^2(60)-2gh, (V@horizontal)^2 =u^2cos^2(60)+2gh, (Results v)^2 = v1^2+v2^2-2v1 * v2cos60. As per me, answer 70(+-)1m/S. Option A. |

Ahmad Mujtaba said: (Feb 10, 2021) | |

v = √(v^2 - 2gh), v = √(100.3^2 - 2*9.8*100). v = 90 m/sec. |

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