To determine the initial speed with which the stone was thrown, we can use the fact that the time taken for the stone to reach its highest point (when it momentarily stops and changes direction) is half of the total time for the stone to complete its trajectory.
Given that the stone returns to the same level 4 seconds later, we can conclude that the stone reaches its highest point after 2 seconds (half of the total time).
Let's denote the initial speed of the stone as v. At the highest point, the stone momentarily stops before falling back down due to gravity. Therefore, its final velocity at the highest point is 0 m/s.
Using the kinematic equation for vertical motion, we can find the initial speed of the stone:
v = u + at
Where: v is the final velocity (0 m/s at the highest point), u is the initial velocity (the speed we want to find), a is the acceleration due to gravity (-10 m/s^2, considering g = 10 m/s^2), t is the time taken to reach the highest point (2 seconds).
Substituting the known values into the equation, we can solve for u:
0 = u + (-10 m/s^2) × (2 s) 0 = u - 20 m/s u = 20 m/s
Therefore, the stone was thrown with an initial speed of 20 m/s.