Initial velocity (u) = 2 m/s
Acceleration (a) = 10 m/s^2
Time taken (t) = 4 s
Let the final velocity be v.
By using the equation,
v = u + at, we get
or, v = 2 + 10 × 4
or, v = 2 + 40
or, v = 42
The final velocity is 42 m/s.
Answer:
6.05 cm
Explanation:
The given equation is
2 aₓ(x-x₀)=( Vₓ²-V₀ₓ²)
The initial head velocity V₀ₓ =11 m/s
The final head velocity Vₓ is 0
The accelerationis given by =1000 m/s²
the stopping distance = x-x₀=?
So we can wind the stopping distance by following formula
2 (-1000)(x-x₀)=[]
x-x₀=6.05* m
=6.05 cm
Answer: Speed = 4 m/s
Explanation:
The parameters given are
Mass M = 60 kg
Height h = 0.8 m
Acceleration due to gravity g= 10 m/s2
Before the man jumps, he will be experiencing potential energy at the top of the table.
P.E = mgh
Substitute all the parameters into the formula
P.E = 60 × 9.8 × 0.8
P.E = 470.4 J
As he jumped from the table and hit the ground, the whole P.E will be converted to kinetic energy according to conservative of energy.
When hitting the ground,
K.E = P.E
Where K.E = 1/2mv^2
Substitute m and 470.4 into the formula
470.4 = 1/2 × 60 × V^2
V^2 = 470.4/30
V^2 = 15.68
V = square root (15.68)
V = 3.959 m/s
Therefore, the speed of the man when hitting the ground is approximately 4 m/s