A stone is thrown vertically upwards from a point that is 12m above the sea. It then falls into the sea below after 3.4 s. Air resistance is negligible. At which speed was the stone released when it was thrown?
Answers
Answered by
0
Explanation:
The equation for a stone in motion would be h(t)=g/2 t²+vt+c, where h(t) is the height of the object in question at any given time t, v is the initial velocity of the object, g is gravity (-9.8 m/s² at sea level), and c is the initial height of the object. So, given:
h(3.4)=0
we have:
0=-4.9(3.4)²+(3.4)v+12
3.4v=44.644
v=13.131 m/s as the original velocity of the stone
Similar questions