A trovey moving on horizontal back of heights of 1.2m strikes a banner at the edge of the back the brass Mass on the trap of the trolley flies of an impact and lead on the graund 2.5m from edge on the boach
It seems like there may be some typos or errors in the question you have provided. However, based on the information given, we can calculate the horizontal distance the brass mass traveled after being struck by the trolley.
Given:
Height of the trolley back (h) = 1.2m
Distance the brass mass landed from the edge of the back (d) = 2.5m
We can use the principle of conservation of energy to find the horizontal distance traveled by the brass mass after being struck by the trolley.
The initial potential energy of the brass mass = mgh, where m is the mass of the brass mass, g is the acceleration due to gravity, and h is the height of the trolley back.
The final kinetic energy of the brass mass = 0.5mv^2, where v is the final velocity of the brass mass when it lands on the ground.
Equating these two energies, we have:
mgh = 0.5mv^2
Solving for v, we get:
gh = 0.5v^2
2gh = v^2
v = sqrt(2gh)
Given that h = 1.2m and g = 9.81 m/s^2, we can calculate the final velocity v of the brass mass when it lands on the ground.
v = sqrt(2*9.81*1.2) = 5.38 m/s
Now, we can calculate the horizontal distance traveled by the brass mass using the formula:
d = vt, where t is the time taken for the brass mass to travel the horizontal distance d.
t = d/v
t = 2.5 / 5.38 = 0.465s
Finally, we can calculate the horizontal distance traveled by the brass mass:
d = vt
d = 5.38 * 0.465 = 2.5m
Therefore, the brass mass traveled a horizontal distance of 2.5 meters after being struck by the trolley.