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A ball is projected from the ground at an angle of 45° with the horizontal surface. It reaches a maximum height of 120 m and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of 30° with the horizontal surface. The maximum height it reaches after the bounce, in meters is ___________.
Correct answer is '30'. Can you explain this answer?
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Verified Answer
A ball is projected from the ground at an angle of 45° with the ho...
Consider the diagram showing the given condition for the ball.
For the angle 45°, the maximum height is calculated as,
The velocity of the ball becomes u' after hitting the ground and it loses half of its kinetic energy at that time, that is,
The maximum height for the angle 30° is calculated below,
= 30 m
For the angle 45°, the maximum height is calculated as,
The velocity of the ball becomes u' after hitting the ground and it loses half of its kinetic energy at that time, that is,
The maximum height for the angle 30° is calculated below,
= 30 m
Most Upvoted Answer
A ball is projected from the ground at an angle of 45° with the ho...
Degrees with the horizontal. The initial speed of the ball is 20 m/s. Neglecting air resistance, find the time it takes for the ball to reach its maximum height and the maximum height reached by the ball.
To find the time it takes for the ball to reach its maximum height, we can use the fact that the vertical component of velocity at the maximum height is zero.
The initial velocity of the ball can be split into its horizontal and vertical components. Since the ball is projected at an angle of 45 degrees, the horizontal and vertical components of velocity are the same.
The initial vertical velocity (Vy) can be found using the formula Vy = V * sin(θ), where V is the initial speed and θ is the angle of projection.
Vy = 20 m/s * sin(45°) = 20 m/s * 0.707 = 14.14 m/s
The time it takes for the ball to reach its maximum height can be found using the formula t = Vy / g, where g is the acceleration due to gravity.
t = 14.14 m/s / 9.8 m/s^2 = 1.44 s
Therefore, it takes 1.44 seconds for the ball to reach its maximum height.
To find the maximum height reached by the ball, we can use the formula H = Vy^2 / (2 * g), where H is the maximum height.
H = (14.14 m/s)^2 / (2 * 9.8 m/s^2) = 100 m
Therefore, the maximum height reached by the ball is 100 meters.
To find the time it takes for the ball to reach its maximum height, we can use the fact that the vertical component of velocity at the maximum height is zero.
The initial velocity of the ball can be split into its horizontal and vertical components. Since the ball is projected at an angle of 45 degrees, the horizontal and vertical components of velocity are the same.
The initial vertical velocity (Vy) can be found using the formula Vy = V * sin(θ), where V is the initial speed and θ is the angle of projection.
Vy = 20 m/s * sin(45°) = 20 m/s * 0.707 = 14.14 m/s
The time it takes for the ball to reach its maximum height can be found using the formula t = Vy / g, where g is the acceleration due to gravity.
t = 14.14 m/s / 9.8 m/s^2 = 1.44 s
Therefore, it takes 1.44 seconds for the ball to reach its maximum height.
To find the maximum height reached by the ball, we can use the formula H = Vy^2 / (2 * g), where H is the maximum height.
H = (14.14 m/s)^2 / (2 * 9.8 m/s^2) = 100 m
Therefore, the maximum height reached by the ball is 100 meters.
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A ball is projected from the ground at an angle of 45° with the horizontal surface. It reaches a maximum height of 120 m and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of 30° with thehorizontal surface. The maximum height it reaches after the bounce, in meters is ___________.Correct answer is '30'. Can you explain this answer?
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A ball is projected from the ground at an angle of 45° with the horizontal surface. It reaches a maximum height of 120 m and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of 30° with thehorizontal surface. The maximum height it reaches after the bounce, in meters is ___________.Correct answer is '30'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A ball is projected from the ground at an angle of 45° with the horizontal surface. It reaches a maximum height of 120 m and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of 30° with thehorizontal surface. The maximum height it reaches after the bounce, in meters is ___________.Correct answer is '30'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball is projected from the ground at an angle of 45° with the horizontal surface. It reaches a maximum height of 120 m and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of 30° with thehorizontal surface. The maximum height it reaches after the bounce, in meters is ___________.Correct answer is '30'. Can you explain this answer?.
A ball is projected from the ground at an angle of 45° with the horizontal surface. It reaches a maximum height of 120 m and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of 30° with thehorizontal surface. The maximum height it reaches after the bounce, in meters is ___________.Correct answer is '30'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A ball is projected from the ground at an angle of 45° with the horizontal surface. It reaches a maximum height of 120 m and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of 30° with thehorizontal surface. The maximum height it reaches after the bounce, in meters is ___________.Correct answer is '30'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball is projected from the ground at an angle of 45° with the horizontal surface. It reaches a maximum height of 120 m and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of 30° with thehorizontal surface. The maximum height it reaches after the bounce, in meters is ___________.Correct answer is '30'. Can you explain this answer?.
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A ball is projected from the ground at an angle of 45° with the horizontal surface. It reaches a maximum height of 120 m and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of 30° with thehorizontal surface. The maximum height it reaches after the bounce, in meters is ___________.Correct answer is '30'. Can you explain this answer?, a detailed solution for A ball is projected from the ground at an angle of 45° with the horizontal surface. It reaches a maximum height of 120 m and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of 30° with thehorizontal surface. The maximum height it reaches after the bounce, in meters is ___________.Correct answer is '30'. Can you explain this answer? has been provided alongside types of A ball is projected from the ground at an angle of 45° with the horizontal surface. It reaches a maximum height of 120 m and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of 30° with thehorizontal surface. The maximum height it reaches after the bounce, in meters is ___________.Correct answer is '30'. Can you explain this answer? theory, EduRev gives you an
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