A mango is dropped and falls freely from rest what are its position and velocity after 1.0s,2.0s, and 3.0s

To find the position and velocity of a mango dropped freely from rest after certain time intervals, we can use the equations of motion. In this case, we are given the time intervals of 1.0s, 2.0s, and 3.0s.

The first equation we can use is the position equation:
s = ut + (1/2)at^2

where:
s = position (distance traveled)
u = initial velocity (0 m/s since it starts from rest)
t = time
a = acceleration (in this case, the acceleration is due to gravity and its value is -9.8 m/s^2)

The second equation we can use is the velocity equation:
v = u + at

Now, let's calculate the position and velocity for each time interval.

1. When t = 1.0s:
Position:
s = (0)(1.0) + (1/2)(-9.8)(1.0^2)
s = 0 - 4.9(1.0)
s = -4.9 meters

Velocity:
v = 0 + (-9.8)(1.0)
v = -9.8 m/s

2. When t = 2.0s:
Position:
s = (0)(2.0) + (1/2)(-9.8)(2.0^2)
s = 0 - 4.9(4.0)
s = -19.6 meters

Velocity:
v = 0 + (-9.8)(2.0)
v = -19.6 m/s

3. When t = 3.0s:
Position:
s = (0)(3.0) + (1/2)(-9.8)(3.0^2)
s = 0 - 4.9(9.0)
s = -44.1 meters

Velocity:
v = 0 + (-9.8)(3.0)
v = -29.4 m/s

Therefore, the position and velocity of the mango after 1.0s, 2.0s, and 3.0s are as follows:

- After 1.0s: Position = -4.9 meters, Velocity = -9.8 m/s
- After 2.0s: Position = -19.6 meters, Velocity = -19.6 m/s
- After 3.0s: Position = -44.1 meters, Velocity = -29.4 m/s

a = -9.81 =-g

v = -9.8 t
h =Hi - Vit - 4.9 t^2 = Hi -4.9 t^2

You do not say the original height, Hi
Vi, initial speed, is 0
so
t ____ a ___ v ___ h

0 ___ -g ___ 0 ___ Hi - 0
1 ___ -g ___-9.8__ Hi - 4.9
2____ -g____-19.6_ Hi - 19.6
3 ___ -g ___-29.4_ Hi - 44.1