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
According to Newton's Second Law of Motion :
The Force acting on an Object is equal to Product of Mass of the Object and Acceleration produced due to the Force.
Force acting = Mass of the Object × Acceleration
Given : Force = 50 newton and Mass of the Object = 10 kg
Substituting the respective values in the Formula, we get :
50 N = 10 kg × Acceleration
Acceleration of the Object = 5 m/s²