Question

5. An object is projected with velocity ū = (aî +bſ) from ground. It has velocity (xi + y) at height h. Value of h is (neglect air resistance) у ü= ai + bj →X o (1) a+ b2 - y2 2g (2) a² - y² 29 b²-y² 29 6² - x² (3) (4) 2g ens​

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

h = (a²+ b²-x²-y²) /(2g)

Explanation:

m = mass of the object.

Total energy of the object at the time of projection

= 1/2 m |ū|² = 1/2 m (a²+b²)

Kinetic Energy of the object at height h

= 1/2 m | x î + y j |² = 1/2 m (x²+y²)

Potential energy of the object at height h

= m g h

Hence applying the law of conservation of energy, we obtain

mgh = 1/2 m (a²+b²) - 1/2 m (x²+y²)

h = (a²+ b²-x²-y²) /(2g)