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A ball thrown vertically upward with a speed of 20 m/sec from the top of a tower returns to the earth in 6 sec. The height of the tower (take g = 10 m/sec 2) –
- a)60 m
- b)55 m
- c)52 m
- d)None
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A ball thrown vertically upward with a speed of 20 m/sec from the top ...
Let after time t1 the ball reached to the topmost point where the velocity will be zero.
So the falling time of the ball is, t2 = 6 – 2 = 4 sec
Now, difference in distance travelled = height of the tower
⇒ height of tower
Shortcut: Net displacement, s = - h (negative sign as the direction of motion is opposite to the displacement)
Most Upvoted Answer
A ball thrown vertically upward with a speed of 20 m/sec from the top ...
To find the height of the tower, we can use the formula for the height of an object in free fall:
h = v0*t + (1/2)*g*t^2
where:
h = height of the object (in this case, the tower)
v0 = initial velocity of the object (20 m/sec)
t = time taken for the object to fall (6 sec)
g = acceleration due to gravity (10 m/sec^2)
Substituting the given values into the formula, we get:
h = (20 m/sec)*(6 sec) + (1/2)*(10 m/sec^2)*(6 sec)^2
h = 120 m + (1/2)*(10 m/sec^2)*(36 sec^2)
h = 120 m + 180 m
h = 300 m
Therefore, the height of the tower is 300 meters.
h = v0*t + (1/2)*g*t^2
where:
h = height of the object (in this case, the tower)
v0 = initial velocity of the object (20 m/sec)
t = time taken for the object to fall (6 sec)
g = acceleration due to gravity (10 m/sec^2)
Substituting the given values into the formula, we get:
h = (20 m/sec)*(6 sec) + (1/2)*(10 m/sec^2)*(6 sec)^2
h = 120 m + (1/2)*(10 m/sec^2)*(36 sec^2)
h = 120 m + 180 m
h = 300 m
Therefore, the height of the tower is 300 meters.
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A ball thrown vertically upward with a speed of 20 m/sec from the top of a tower returns to the earth in 6 sec. The height of the tower (take g = 10 m/sec 2) –a)60 mb)55 mc)52 md)NoneCorrect answer is option 'A'. Can you explain this answer?
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