Answer:
The horizontal distance traveled when the stone returns to its original height = 220.81 m
Explanation:
Considering vertical motion of catapult:-
At maximum height,
Initial velocity, u = 50 sin30 = 25 m/s
Acceleration , a = -9.81 m/s²
Final velocity, v = 0 m/s
We have equation of motion v = u + at
Substituting
v = u + at
0 = 25 - 9.81 x t
t = 2.55 s
Time of flight = 2 x Time to reach maximum height = 2 x 2.55 = 3.1 s
Considering horizontal motion of catapult:-
Initial velocity, u = 50 cos30 = 43.30 m/s
Acceleration , a = 0 m/s²
Time, t = 5.10 s
We have equation of motion s= ut + 0.5 at²
Substituting
s= ut + 0.5 at²
s = 43.30 x 5.10 + 0.5 x 0 x 5.10²
s = 220.81 m
The horizontal distance traveled when the stone returns to its original height = 220.81 m
