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From a tower of height 100m , a ball is dropped at same instant another ball is thrown vertically upward with velocity 20m/s from base of the tower in same line after what time both balls will hit each other also find the height when they hit each other.?
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From a tower of height 100m , a ball is dropped at same instant anothe...
Analysis:
To solve this problem, we need to consider the motion of both balls separately and find the time at which their paths intersect. Let's break down the problem into steps:
Step 1: Analyzing the downward motion of the ball:
The ball dropped from a height of 100m falls under the influence of gravity. We can use the equation of motion to find the time it takes to reach the ground.
Using the equation:
h = ut + (1/2)gt^2
Where:
h = height (100m),
u = initial velocity (0 m/s, as it is dropped),
g = acceleration due to gravity (-9.8 m/s^2),
t = time.
Plugging in the values, we get:
100 = 0 + (1/2)(-9.8)t^2
100 = -4.9t^2
Simplifying the equation, we find:
t^2 = 100/4.9
t^2 = 20.41
t ≈ 4.52 seconds
So, the ball takes approximately 4.52 seconds to reach the ground.
Step 2: Analyzing the upward motion of the ball:
The other ball is thrown vertically upwards with an initial velocity of 20 m/s. We can use the equation of motion to find the time it takes to reach the same height as the dropped ball.
Using the equation:
h = ut + (1/2)gt^2
Where:
h = height (100m),
u = initial velocity (20 m/s),
g = acceleration due to gravity (-9.8 m/s^2),
t = time.
Plugging in the values, we get:
100 = 20t + (1/2)(-9.8)t^2
100 = 20t - 4.9t^2
Rearranging the equation, we get:
4.9t^2 - 20t + 100 = 0
Solving this quadratic equation using the quadratic formula, we find:
t ≈ 4.92 seconds
So, the ball takes approximately 4.92 seconds to reach a height of 100m.
Step 3: Finding the time and height of intersection:
To find the time at which both balls intersect, we need to consider the time of flight for the upward ball. As the upward ball will take a longer time to reach the same height as the dropped ball, we can use the time of flight of the upward ball as the common time for intersection.
The time of flight for the upward ball can be calculated using the equation:
t_flight = 2u/g
Where:
u = initial velocity (20 m/s),
g = acceleration due to gravity (-9.8 m/s^2).
Plugging in the values, we get:
t_flight = 2(20)/(-9.8)
t_flight ≈ -4.08 seconds
Since time cannot be negative, we can ignore the negative value.
Therefore, both balls will hit each other after approximately 4.08 seconds.
To find the height at which they hit each other, we can substitute the time of flight into the equation of motion for the dropped ball.
Using the equation:
h = ut + (1/2)gt^2
Where:
h =
To solve this problem, we need to consider the motion of both balls separately and find the time at which their paths intersect. Let's break down the problem into steps:
Step 1: Analyzing the downward motion of the ball:
The ball dropped from a height of 100m falls under the influence of gravity. We can use the equation of motion to find the time it takes to reach the ground.
Using the equation:
h = ut + (1/2)gt^2
Where:
h = height (100m),
u = initial velocity (0 m/s, as it is dropped),
g = acceleration due to gravity (-9.8 m/s^2),
t = time.
Plugging in the values, we get:
100 = 0 + (1/2)(-9.8)t^2
100 = -4.9t^2
Simplifying the equation, we find:
t^2 = 100/4.9
t^2 = 20.41
t ≈ 4.52 seconds
So, the ball takes approximately 4.52 seconds to reach the ground.
Step 2: Analyzing the upward motion of the ball:
The other ball is thrown vertically upwards with an initial velocity of 20 m/s. We can use the equation of motion to find the time it takes to reach the same height as the dropped ball.
Using the equation:
h = ut + (1/2)gt^2
Where:
h = height (100m),
u = initial velocity (20 m/s),
g = acceleration due to gravity (-9.8 m/s^2),
t = time.
Plugging in the values, we get:
100 = 20t + (1/2)(-9.8)t^2
100 = 20t - 4.9t^2
Rearranging the equation, we get:
4.9t^2 - 20t + 100 = 0
Solving this quadratic equation using the quadratic formula, we find:
t ≈ 4.92 seconds
So, the ball takes approximately 4.92 seconds to reach a height of 100m.
Step 3: Finding the time and height of intersection:
To find the time at which both balls intersect, we need to consider the time of flight for the upward ball. As the upward ball will take a longer time to reach the same height as the dropped ball, we can use the time of flight of the upward ball as the common time for intersection.
The time of flight for the upward ball can be calculated using the equation:
t_flight = 2u/g
Where:
u = initial velocity (20 m/s),
g = acceleration due to gravity (-9.8 m/s^2).
Plugging in the values, we get:
t_flight = 2(20)/(-9.8)
t_flight ≈ -4.08 seconds
Since time cannot be negative, we can ignore the negative value.
Therefore, both balls will hit each other after approximately 4.08 seconds.
To find the height at which they hit each other, we can substitute the time of flight into the equation of motion for the dropped ball.
Using the equation:
h = ut + (1/2)gt^2
Where:
h =
Community Answer
From a tower of height 100m , a ball is dropped at same instant anothe...
80.4 m
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From a tower of height 100m , a ball is dropped at same instant another ball is thrown vertically upward with velocity 20m/s from base of the tower in same line after what time both balls will hit each other also find the height when they hit each other.?
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From a tower of height 100m , a ball is dropped at same instant another ball is thrown vertically upward with velocity 20m/s from base of the tower in same line after what time both balls will hit each other also find the height when they hit each other.? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about From a tower of height 100m , a ball is dropped at same instant another ball is thrown vertically upward with velocity 20m/s from base of the tower in same line after what time both balls will hit each other also find the height when they hit each other.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for From a tower of height 100m , a ball is dropped at same instant another ball is thrown vertically upward with velocity 20m/s from base of the tower in same line after what time both balls will hit each other also find the height when they hit each other.?.
From a tower of height 100m , a ball is dropped at same instant another ball is thrown vertically upward with velocity 20m/s from base of the tower in same line after what time both balls will hit each other also find the height when they hit each other.? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about From a tower of height 100m , a ball is dropped at same instant another ball is thrown vertically upward with velocity 20m/s from base of the tower in same line after what time both balls will hit each other also find the height when they hit each other.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for From a tower of height 100m , a ball is dropped at same instant another ball is thrown vertically upward with velocity 20m/s from base of the tower in same line after what time both balls will hit each other also find the height when they hit each other.?.
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