Tangential and Normal Accelerations of a Stone

Given:
- Initial velocity (u) of the stone = 15 m/s
- Time (t) = 1 second

To determine the tangential and normal accelerations of the stone after 1 second:

Step 1: Find the initial horizontal velocity (ux) and initial vertical velocity (uy):
- Since the stone is thrown horizontally, the initial horizontal velocity (ux) is equal to the initial velocity (u) = 15 m/s.
- The initial vertical velocity (uy) is 0 m/s as the stone is not thrown vertically.

Step 2: Calculate the horizontal displacement (Sx) and vertical displacement (Sy):
- The horizontal displacement (Sx) can be calculated using the formula: Sx = ux * t, where t is the time.
- Sx = 15 m/s * 1 s = 15 m
- The vertical displacement (Sy) can be calculated using the formula: Sy = (1/2) * g * t^2, where g is the acceleration due to gravity.
- Sy = (1/2) * 9.8 m/s^2 * (1 s)^2 = 4.9 m

Step 3: Determine the tangential acceleration (at) and normal acceleration (an):
- Tangential acceleration (at) is the rate of change of tangential velocity, which is constant as the stone is moving horizontally.
- at = 0 m/s^2 (since there is no change in horizontal velocity)
- Normal acceleration (an) is the rate of change of vertical velocity, which is equal to the acceleration due to gravity (g) as the stone falls freely.
- an = g = 9.8 m/s^2

Conclusion:
After 1 second of the stone's motion, the tangential acceleration (at) is 0 m/s^2, indicating no change in horizontal velocity. The normal acceleration (an) is equal to the acceleration due to gravity (9.8 m/s^2), representing the stone's vertical acceleration as it falls freely.

An=gcosa=39/√13
at=gsina=2/√13(t=tangential)