Question

A ball is projected horizontally with a velocity of $$ 5ms^{-1} $$ from the top of a tower of height 10 m. When it reaches the ground ( $$ g = 10 ms^{-2} $$ )

Answer 1

Correct option is A. Vertical component of its velocity is $$ 15ms^{-1} $$
$$ V = \sqrt{2gh} = \sqrt{2 \times 10 \times 10}$$
$$ Vy = 10\sqrt{2} $$
$$ Vx = 5 m/s $$ same
$$ V = \sqrt{Vx^{2} + Vy^{2}} $$
$$ = \sqrt{200 + 25} $$
$$ = \sqrt{225} $$
$$ = 15 $$