Answer:
The initial velocity is 50 m/s.
(C) is correct option.
Explanation:
Given that,
Time = 10 sec
For first half,
We need to calculate the height
Using equation of motion
....(I)
For second half,
We need to calculate the time
Using equation of motion
Put the value of h from equation (I)
According to question,
Put the value of t₁ and t₂
Here, g = 10
The initial velocity is
Hence, The initial velocity is 50 m/s.
Answer:
66.4 N
Explanation:
From Newton's second law, <em>F </em>=<em> ma</em>
where <em>F</em> is the force, <em>m</em> is the mass and <em>a</em> is the acceleration.
Because the object has acceleration in two directions and the mass is constant, the force will be in two directions. The component of the forces are:
The magnitude of the resultant force is given by
Answer:
K.E₂ = mg(h - 2R)
Explanation:
The diagram of the car at the top of the loop is given below. Considering the initial position of the car and the final position as the top of the loop. We apply law of conservation of energy:
K.E₁ + P.E₁ = K.E₂ + P.E₂
where,
K.E₁ = Initial Kinetic Energy = (1/2)mv² = (1/2)m(0 m/s)² = 0 (car initially at rest)
P.E₁ = Initial Potential Energy = mgh
K.E₂ = Final Kinetic Energy at the top of the loop = ?
P.E₂ = Final Potential Energy = mg(2R) (since, the height at top of loop is 2R)
Therefore,
0 + mgh = K.E₂ + mg(2R)
<u>K.E₂ = mg(h - 2R)</u>
