Answer:
the rates of rock formation are similar. i could be wrong tho.....
Explanation:
Answer:
10.4 m/s
Explanation:
First, find the time it takes for the projectile to fall 6 m.
Given:
y₀ = 6 m
y = 0 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: t
y = y₀ + v₀ t + ½ at²
(0 m) = (6 m) + (0 m/s) t + ½ (-9.8 m/s²) t²
t = 1.11 s
Now find the horizontal position of the target after that time:
Given:
x₀ = 6 m
v₀ = 5 m/s
a = 0 m/s²
t = 1.11 s
Find: x
x = x₀ + v₀ t + ½ at²
x = (6 m) + (5 m/s) (1.11 s) + ½ (0 m/s²) (1.11 s)²
x = 11.5 m
Finally, find the launch velocity needed to travel that distance in that time.
Given:
x₀ = 0 m
x = 11.5 m
t = 1.11 s
a = 0 m/s²
Find: v₀
(11.5 m) = (0 m) + v₀ (1.11 s) + ½ (0 m/s²) (1.11 s)²
v₀ = 10.4 m/s
I'm not sure if this is correct but it's what I'll do
This is free-fall problem.
Stone A is thrown upward, at the point it falls down to the place where it was thrown, the velocity is -15m/s.
Now I choose the bridge is the origin. From the bridge, stone A and B fall the same distance which means Ya = Yb ( vertical distance )
Ya = Vo(t + 2) + 1/2a(t+2)^2
= -15(t + 2) + 1/2(9.8)(t^2 + 4t + 4)
= -15t - 30 + 4.5(t^2 + 4t + 4)
= -15t - 30 + 4.5t^2 + 18t + 18
= 4.5t^2 +3t - 12
Yb = Vo(t) + 1/2a(t)^2
= 0 + 4.5t^2
4.5t^2 = 4.5t^2 +3t - 12
0 = 3t - 12
4 = t
Time for Stone B is 4s
Time for Stone A is 6s
