Question

A stone is projected horizontally
with a velocity 19.6ms from a
tower of height 50m. Its velocity
two seconds after projection is
(g = 9.8ms)
9.8ms 1​

Answer 1

Given:

A stone is projected horizontally with a velocity 19.6m/s from a tower of height 50m.

To find:

Velocity after 2 seconds after projection ?

Calculation:

Y component velocity after 2 secs:

$\therefore \: v_{y} = u_{y} + gt$

$\implies\: v_{y}= 0 + (9.8)(2)$

$\implies\: v_{y}=19.6 \: m {s}^{ - 1}$

X component of velocity after 2 secs:

• The x component of velocity will remain same as the initial velocity (19.6 m/s) as there is no acceleration along x axis.

$\therefore \: v_{x} = 19.6 \: m {s}^{ - 1}$

Therefore, total velocity:

$\therefore \: v_{net} = \sqrt{ {(v_{x})}^{2} + {(v_{y})}^{2} }$

$\implies \: v_{net} = \sqrt{ {(19.6)}^{2} + {(19.6)}^{2} }$

$\implies \: v_{net} = 19.6 \sqrt{2}$

$\implies \: v_{net} = 27.4 \: m {s}^{ - 1}$

So, net velocity is 27.4 m/s .