A stone is thrown vertically upward at 15 m/s. Neglect friction due to the air. What is the stone's velocity after 2 seconds?
V =Vo -gt^2/2
g =9.8 m/s^2
V = Vo - g*t = 15 - 19.6 = -4.6 m/s
(minus sign indicates it is coming back down)
Yeroo kana nurraa yeroo ammaa kana
To find the stone's velocity after 2 seconds, we can use the equation of motion for an object in free fall.
The equation of motion for an object in free fall is given by:
v = u + gt
Where:
v is the final velocity
u is the initial velocity
g is the acceleration due to gravity (approximated as 9.8 m/s^2)
t is the time
In this case, the stone is initially thrown vertically upward with an initial velocity of 15 m/s. Since the acceleration due to gravity acts in the opposite direction, we take it as negative (-9.8 m/s^2).
Substituting the given values into the equation, we have:
v = 15 m/s + (-9.8 m/s^2) * 2 s
Calculating the expression:
v = 15 m/s - 19.6 m/s
v ≈ -4.6 m/s
Therefore, the stone's velocity after 2 seconds is approximately -4.6 m/s. The negative sign indicates that the stone is moving downward.