438,155 views
41 votes
41 votes
a body is thrown up from a wall of height 2.2m with a speed of 10ms^-1 the distance travelled in the last second of its motion

User SondreB
by
2.6k points

1 Answer

7 votes
7 votes

Answer:

To solve this problem, we need to use the equations of motion for an object undergoing uniformly accelerated motion. Specifically, we can use the equation for the displacement of an object in the vertical direction:

y = y0 + v0t + 1/2at^2

where y is the displacement of the object, y0 is the initial displacement (in this case, the height of the wall), v0 is the initial velocity (in this case, 10 m/s), t is the time elapsed, and a is the acceleration (in this case, the acceleration due to gravity, which is approximately 9.81 m/s^2).

To find the distance travelled in the last second of motion, we need to find the displacement of the object at t = 1 s, and subtract the displacement at t = 0 s (which is the initial displacement). Substituting these values into the equation above and solving, we get:

y1 = 2.2m + 10m/s * 1s + 1/2 * 9.81m/s^2 * 1s^2 = 12.61 m

y0 = 2.2m

The distance travelled in the last second of motion is therefore y1 - y0 = 12.61 m - 2.2 m = 10.41 m.

Step-by-step explanation:

User Jens Roland
by
2.9k points