Question

c)-20 24. A stone thrown from the top of a 50 m tall building is given an initial velocity of 20.0 m/s straight upward. Determine the velocity in m/sec when the stone returns to the height from which it was thrown. g = 9.8 m/sec2 b) -15.0 a) -20.0 d) -25.0 c) -30.0 20 A hall starts from rest and accelerates at 0.500 m/s2 while moving down an inclined plane 9.00 m​

Answer 1

Answer

The velocity is -20m/s at 50 m

Explanation:

Stone thrown from top of the building of height ( $x_0=$)50m,

initial velocity(u) = 20m/s,

The maximum height the stone can reach can be found from $v^2-u^2=2a(x-x_0)$, $a=-10m/s^2$

at maximum height v=0, hence from the equation

$-u^2=-20(x-50)$

putting the values we get

⇒x=70m

The energy at height 70 m=$mgh=m*9.8*70=686m$

When the stone reaches from the height it was thrown that is 50 m, the energy of the stone is $E=\frac{1}{2}mv^2+mgx_0= \frac{1}{2}mv^2+490m$,

According to conservation of energy

$\frac{1}{2}mv^2+490m=686m$

⇒$v=\sqrt{392}=19.7$≈20m/s

since the current motion is downward direction of velocity is opposite from how its thrown, hence the velocity is -20m/s.

Answer 2

Answer:

velocity is 20m/s

Explanation: