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A ball is horizontally from the top of tower of height 100m with a velocity of 5m/s. Calculate the time taken by the ball to reach the ground. Take g =9.8m/ square ?
Most Upvoted Answer
A ball is horizontally from the top of tower of height 100m with a vel...
Tower Height
The tower has a height of 100m.
Initial Velocity
The ball is launched horizontally from the top of the tower with an initial velocity of 5m/s.
Acceleration due to Gravity
The acceleration due to gravity is given as g = 9.8m/s^2.
Time to Reach the Ground
To calculate the time taken by the ball to reach the ground, we can use the equation of motion:
h = ut + (1/2)gt^2
Where:
h = height of the tower (100m)
u = initial velocity (5m/s)
t = time taken
Substituting Values
Plugging in the given values into the equation, we get:
100 = 5t + (1/2)(9.8)t^2
Simplifying the Equation
To solve the equation, we can rearrange it to form a quadratic equation:
(1/2)(9.8)t^2 + 5t - 100 = 0
Using Quadratic Formula
Applying the quadratic formula:
t = (-b ± √(b^2 - 4ac))/(2a)
Where:
a = 1/2(9.8)
b = 5
c = -100
Calculating Time
Plugging in the values into the quadratic formula:
t = (-5 ± √(5^2 - 4(1/2)(9.8)(-100)))/(2(1/2)(9.8))
Simplifying the equation further:
t = (-5 ± √(25 + 1960))/(9.8)
t = (-5 ± √1985)/(9.8)
Since we are dealing with a physical scenario, the negative value of time does not make sense. Therefore, we take the positive value:
t = (-5 + √1985)/(9.8)
Using a calculator to evaluate the square root and perform the addition:
t ≈ 4.04 seconds
Conclusion
The ball will take approximately 4.04 seconds to reach the ground when launched horizontally from the top of a tower with a velocity of 5m/s.
The tower has a height of 100m.
Initial Velocity
The ball is launched horizontally from the top of the tower with an initial velocity of 5m/s.
Acceleration due to Gravity
The acceleration due to gravity is given as g = 9.8m/s^2.
Time to Reach the Ground
To calculate the time taken by the ball to reach the ground, we can use the equation of motion:
h = ut + (1/2)gt^2
Where:
h = height of the tower (100m)
u = initial velocity (5m/s)
t = time taken
Substituting Values
Plugging in the given values into the equation, we get:
100 = 5t + (1/2)(9.8)t^2
Simplifying the Equation
To solve the equation, we can rearrange it to form a quadratic equation:
(1/2)(9.8)t^2 + 5t - 100 = 0
Using Quadratic Formula
Applying the quadratic formula:
t = (-b ± √(b^2 - 4ac))/(2a)
Where:
a = 1/2(9.8)
b = 5
c = -100
Calculating Time
Plugging in the values into the quadratic formula:
t = (-5 ± √(5^2 - 4(1/2)(9.8)(-100)))/(2(1/2)(9.8))
Simplifying the equation further:
t = (-5 ± √(25 + 1960))/(9.8)
t = (-5 ± √1985)/(9.8)
Since we are dealing with a physical scenario, the negative value of time does not make sense. Therefore, we take the positive value:
t = (-5 + √1985)/(9.8)
Using a calculator to evaluate the square root and perform the addition:
t ≈ 4.04 seconds
Conclusion
The ball will take approximately 4.04 seconds to reach the ground when launched horizontally from the top of a tower with a velocity of 5m/s.
Community Answer
A ball is horizontally from the top of tower of height 100m with a vel...
Using s=ut+1/2 at^2 formula we can find answer.
s=200m
u=5m/s
a=9.8m/s^2
we will get a quadratic equation, by solving you will get answer as 4.03 approximately is 4seconds
s=200m
u=5m/s
a=9.8m/s^2
we will get a quadratic equation, by solving you will get answer as 4.03 approximately is 4seconds
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A ball is horizontally from the top of tower of height 100m with a velocity of 5m/s. Calculate the time taken by the ball to reach the ground. Take g =9.8m/ square ?
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A ball is horizontally from the top of tower of height 100m with a velocity of 5m/s. Calculate the time taken by the ball to reach the ground. Take g =9.8m/ square ? 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 A ball is horizontally from the top of tower of height 100m with a velocity of 5m/s. Calculate the time taken by the ball to reach the ground. Take g =9.8m/ square ? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball is horizontally from the top of tower of height 100m with a velocity of 5m/s. Calculate the time taken by the ball to reach the ground. Take g =9.8m/ square ?.
A ball is horizontally from the top of tower of height 100m with a velocity of 5m/s. Calculate the time taken by the ball to reach the ground. Take g =9.8m/ square ? 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 A ball is horizontally from the top of tower of height 100m with a velocity of 5m/s. Calculate the time taken by the ball to reach the ground. Take g =9.8m/ square ? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball is horizontally from the top of tower of height 100m with a velocity of 5m/s. Calculate the time taken by the ball to reach the ground. Take g =9.8m/ square ?.
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