A girl throw a ball with initial velocity V at an inclination of 45° .The ball strikes the smooth vertical wall at a horizontal distance d from the girl & after rebounding returns to her hand .What is the coefficient oof resitution between wall & the ball
To solve this problem, we can use the principle of conservation of energy.
Let's break down the problem step by step.
Step 1: Calculate the maximum height reached by the ball.
Given that the initial velocity (V) makes an angle of 45° with the ground, we can split the initial velocity into its horizontal and vertical components.
The horizontal component of velocity (Vx) = V * cos(45°) = V * √2 / 2
The vertical component of velocity (Vy) = V * sin(45°) = V * √2 / 2
At the maximum height, the vertical component of velocity becomes zero. So we can use the equation of motion: Vy^2 = Uy^2 - 2 * g * h, where Uy is the initial vertical velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the maximum height reached.
Since the ball returns to the girl's hand, the final vertical velocity (Vfy) is equal to the initial vertical velocity (Vy). So we have Vy = -Uy (negative sign because the direction is opposite).
Plugging in the values, we get:
0 = (V * √2 / 2)^2 - 2 * 9.8 * h
Simplifying the equation, we get:
h = V^2 / (8 * g)
Step 2: Calculate the time taken to reach the maximum height.
We can use the equation of motion: Vy = Uy + g * t, where t is the time taken.
Plugging in the values, we get:
0 = (V * √2 / 2) - 9.8 * t
Solving for t, we get:
t = V / (2 * 9.8)
Step 3: Calculate the horizontal distance covered by the ball.
Since the ball strikes the smooth vertical wall at a horizontal distance (d) from the girl, we can calculate the time taken for the ball to reach the wall.
The time taken to reach the wall (t1) is equal to twice the time taken to reach the maximum height (t). So we have:
t1 = 2 * t
The horizontal distance covered (d) is given by the equation: d = Vx * t1
Plugging in the values, we get:
d = (V * √2 / 2) * (2 * V / (2 * 9.8))
Simplifying the equation, we get:
d = V^2 / (9.8 * √2)
Step 4: Calculate the coefficient of restitution (e).
The coefficient of restitution (e) is given by the equation: e = (Vf - Vi) / (Ui - Uf), where Vi and Uf are the initial and final velocities before and after the collision, and Vf and Ui are the final and initial velocities after and before the collision.
In this case, the initial velocity before the collision (Vi) is V, and the final velocity after the collision (Vf) is -V. The initial vertical velocity before the collision (Ui) is Vy, and the final vertical velocity after the collision (Uf) is -Vy.
Plugging in the values, we get:
e = (-V - V) / (Vy - (-Vy))
Simplifying the equation, we get:
e = -2V / (2 * Vy)
Since Vy = V * √2 / 2, we get:
e = -2V / (2 * V * √2 / 2)
Simplifying the equation further, we get:
e = -2V / (V * √2)
e = -√2
Therefore, the coefficient of restitution between the wall and the ball is -√2.