Question

A stone thrown with the velocity V₀ = 14 m/s at an angle 45° to the horizontal, dropped to the ground at a distance 'S' from the point where it was thrown. From what height should the stone be thrown in horizontal direction with the same initial velocity so that it fall at the same spot –

• A. 14.2 m
• B. 16.9 m
• C. 10.0 m
• D. 9.6 m

Answer 1

Answer: (C)

10.0 m

Answer 2

$\frac{u^{2}\sin 2\theta}{g}$ = $u\sqrt{\frac{2h}{g}}$

$\frac{(14)^{2}\sin 2(45)}{9.8}$ = $14\sqrt{\frac{2(h)}{9.8}}$

h = 10 m