Question

A stone is thrown horizontally with velocity 30 m/s from the top of a tower height of 60m. The velocity with which it hits ground is (in m/s)
1) 0
2) 20
3) 10 rout 21
4) 20 rout 3
say answer with solution please ​

Answer 1

Explanation:

Given,

Vertical velocity is zero

Horizontal velocity is v=30m/s

Height, h=60m

Time period t= √2h/g = √2×60/9.81 = 8.64962423 sec

Range, R=vt=30×8.64962423= 259.488727m

Answer 2

Given :

Initial velocity of stone = 30m/s

Height of the tower = 60m

To Find :

Final velocity of stone with which it hits ground.

Solution :

• Horizontal component of velocity remains constant throughout the motion.
• Vertical component of velocity changes with time. At any time t, vertical velocity of object is given by u' = √2gH

» Horizontal component = u

» Vertical component = u'

Final velocity of stone is given by;

$\sf:\implies\:v=\sqrt{u^2+u'^2}$

$\sf:\implies\:v=\sqrt{u^2+(\sqrt{2gH})^2}$

$\sf:\implies\:v=\sqrt{u^2+2gH}$

$\sf:\implies\:v=\sqrt{(30)^2+2(10)(60)}$

$\sf:\implies\:v=\sqrt{900+1200}$

$\sf:\implies\:v=\sqrt{2100}$

$\sf:\implies\:v=\sqrt{21\times 100}$

$:\implies\:\underline{\boxed{\bf{\gray{v=10\sqrt{21}\:ms^{-1}}}}}$