A body is projected vertically upward with a speed of 40 m/s. Find the distance travelled by the body in the last sec of the upward journey taking g = 9.8m/s
in the last second, the velocity goes from 9.8 m/s to zero
the average velocity is 4.9 m/s for one sec
To find the distance traveled by the body in the last second of its upward journey, we need to calculate the height it reaches in that last second.
The equation we can use to calculate the height reached is:
h = ut - 0.5gt^2
where:
h = height reached
u = initial velocity (40 m/s)
g = acceleration due to gravity (9.8 m/s^2)
t = time (in seconds)
Since we want to find the height reached in the last second of the upward journey, we can substitute t = 1 second into the equation:
h = (40 * 1) - 0.5 * 9.8 * (1^2)
h = 40 - 0.5 * 9.8 * 1
h = 40 - 0.5 * 9.8
h = 40 - 4.9
h = 35.1 meters
Therefore, the body reaches a height of 35.1 meters in the last second of its upward journey.
Note: The distance traveled by the body in the last second of the upward journey is equal to the height reached in that last second.