If the escape velocity of a body on Earth is 11.2 km/s, the escape velocity of the body thrown at an angle 45° with the horizontal will be

Correct Option is (B) 125m \(H_{max} = \frac{u^2}{2g}\) \(H_{max} = \frac{50 \times 50}{2 \times 10}\) \(H_{max} = 125 m\)

Correct Answer - (i) 80 m (ii) 8 sec (iii) 320 m (iv) 40 m/s (v) 1 : 1 (vi)`y=x-(x^(2))/(320)m` (vii) 80 m (viii) 60 m

Correct Answer - (i)`theta=45^(@)` (ii)`R=(u^(2))/(g),H=(u^(2))/(4g)` (iii) 1 : 1

Correct Answer - (a)`("mu"^(3)sin^(2)thetacostheta)/(2g)(b)(2"mu"^(3)sin^(2)thetacostheta)/(g)`

Correct option is: (B) \(2v_e\) Escape Velocity \(v_e = \sqrt {2gRe}\) \(\frac{V_{e1}}{V_{e2}} \propto \sqrt \frac{4R_1}{R_1}\) \(V_{escape \ 2} = 2 \ V_{escape \ 1}\)

Correct option is: (A) 11.2 km/s Escape velocity does not depend on the angle of projection. \(V_{escape} = \sqrt {2gRe} \)

C) A + d = i1 + i2 

Correct option is (B) 30° Let the angle be x. Then its complementary angle is 2x. \(\therefore\) x+2x = \(90^\circ\) \(\Rightarrow\) 3x = \(90^\circ\) \(\Rightarrow\) x = \(\frac{90^\circ}3=30^\circ\) Hence, the required angle is \(30^\circ.\)