Answer:
2.65m/s
Explanation:
Using the equation of motion:
v² = u²+2a∆S where
v is the final velocity
u is the initial velocity
∆S is the change in distance
a is the acceleration
Given
u = 0m/s
a = 9.8m/s²
∆S = 1.3-0.943
∆S = 0.357m
Substituting the given parameters into the formula
v² = 0²+2(9.8)(0.357)
v² = 0+6.9972
v² = 6.9972
v=√6.9972
v = 2.65m/s
Hence the velocity at which it hit the ground is 2.65m/s
"The <span>ground is positively charged and the clouds are negatively charged " is the statement among the statements given in the question that </span><span>best explains the movement of electric current from the clouds to the ground during a lightning storm. The correct option among all the options that are given in the question is the third option or option "C". </span>
v = v₀ + at
v = final speed, v₀ = initial speed, a = acceleration, t = elapsed time
Given values:
v₀ = 0m/s (starts from rest), a = 9.81m/s², t = 3s
Plug in and solve for v:
v = 0 + 9.81(3)
v = 29.4m/s
The answer is going to be leaves.