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A ball strikes a smooth horizontal ground at an angle of 45° with the vertical. What cannot be the possible angle of its velocity with the vertical after the collision. (Assume e <=1).
- a)45°
- b)30°
- c)53°
- d)60°
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A ball strikes a smooth horizontal ground at an angle of 45° with ...
The x component will remain same while the y component will become ‘e’ times.
Now to find the new angle
tanα = 1/e
Since, e<=1 tanα >= 1 so α >= initial angle
Most Upvoted Answer
A ball strikes a smooth horizontal ground at an angle of 45° with ...
Conserve momentum in only in y axis
& then apply coefficient of restitution eq.
& then apply coefficient of restitution eq.
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A ball strikes a smooth horizontal ground at an angle of 45° with ...
Degrees with the horizontal. The initial velocity of the ball is 20 m/s. Determine the time it takes for the ball to hit the ground and the horizontal distance it travels before hitting the ground.
To solve this problem, we can break the initial velocity of the ball into its horizontal and vertical components. Since the angle of projection is 45 degrees, the horizontal and vertical components will be equal.
Horizontal component of velocity (Vx) = initial velocity (V) * cos(angle)
Vx = 20 m/s * cos(45 degrees)
Vx = 20 m/s * 0.7071
Vx ≈ 14.14 m/s
Vertical component of velocity (Vy) = initial velocity (V) * sin(angle)
Vy = 20 m/s * sin(45 degrees)
Vy = 20 m/s * 0.7071
Vy ≈ 14.14 m/s
Now, we can analyze the vertical motion of the ball. The ball is thrown upwards, reaches its highest point, and then falls back down. At the highest point, the vertical velocity becomes zero.
Using the equation of motion for vertical motion:
Vy = initial vertical velocity (Uy) + acceleration due to gravity (g) * time (t)
0 = 14.14 m/s - 9.8 m/s^2 * t
Solving for t:
t = 14.14 m/s / 9.8 m/s^2
t ≈ 1.44 seconds
Now, we can determine the horizontal distance traveled by the ball using the horizontal component of velocity (Vx) and the time of flight (t).
Horizontal distance (D) = horizontal velocity (Vx) * time (t)
D = 14.14 m/s * 1.44 seconds
D ≈ 20.38 meters
Therefore, it takes approximately 1.44 seconds for the ball to hit the ground, and it travels approximately 20.38 meters horizontally before hitting the ground.
To solve this problem, we can break the initial velocity of the ball into its horizontal and vertical components. Since the angle of projection is 45 degrees, the horizontal and vertical components will be equal.
Horizontal component of velocity (Vx) = initial velocity (V) * cos(angle)
Vx = 20 m/s * cos(45 degrees)
Vx = 20 m/s * 0.7071
Vx ≈ 14.14 m/s
Vertical component of velocity (Vy) = initial velocity (V) * sin(angle)
Vy = 20 m/s * sin(45 degrees)
Vy = 20 m/s * 0.7071
Vy ≈ 14.14 m/s
Now, we can analyze the vertical motion of the ball. The ball is thrown upwards, reaches its highest point, and then falls back down. At the highest point, the vertical velocity becomes zero.
Using the equation of motion for vertical motion:
Vy = initial vertical velocity (Uy) + acceleration due to gravity (g) * time (t)
0 = 14.14 m/s - 9.8 m/s^2 * t
Solving for t:
t = 14.14 m/s / 9.8 m/s^2
t ≈ 1.44 seconds
Now, we can determine the horizontal distance traveled by the ball using the horizontal component of velocity (Vx) and the time of flight (t).
Horizontal distance (D) = horizontal velocity (Vx) * time (t)
D = 14.14 m/s * 1.44 seconds
D ≈ 20.38 meters
Therefore, it takes approximately 1.44 seconds for the ball to hit the ground, and it travels approximately 20.38 meters horizontally before hitting the ground.
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A ball strikes a smooth horizontal ground at an angle of 45° with the vertical. What cannot be the possible angle of its velocity with the vertical after the collision. (Assume e <=1).a)45°b)30°c)53°d)60°Correct answer is option 'B'. Can you explain this answer?
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