Question

3. A stone of mass 1kg is thrown at 10m/s upward making an angle of 37° with the horizontal
from a building that is 20m high. Using the law of conservation of energy, calculate the
speed when the stone hits the ground.​

Answer 1

Given:

m = 1Kg

v = 10m/s

h = 20m

To Find:

Speed when the stone hits the ground using law of conservation of energy

Solution:

By the law of energy conservation :

$\frac{mv^2}{2}+mgh=\frac{mu^2}{2}$    

Where, v is initial speed, uu is final speed, hh is building height

When the stone falls back down and reaches the horizontal position again, it will have the same angle as at launch: 37°

Horizontal component of velocity:

$v_x=u_x=10cos37°\\=8 m/s$

this does not change

Vertical component of initial velocity:

$v_y =10sin37°=6 m/s$

So:

$u_y=\sqrt{v^2_y+2gh}=\sqrt{6^2+2\cdot9.8\cdot20$

$20.7\frac{m}{s}$

The final speed:

u = $\sqrt{8^2+20.7^2}$

= 22.2 $\frac{m}{s}$