A stone is thrown vertically upward and it returns within 40 seconds. Calculate the maximum height achieved by the stone and the velocity by which it but the ground.
Answers
Answer:
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Explanation:
Soln,
A stone is thrown vertically upward, that means
its final velocity (v) = 0
g = 10 m/s2
time taken to return (t) = 40
we know
u = v + gt
or, u = 0 + 10x 40
or, u = 40 m/s
Now,
s = u + v / 2 x t
or, s = 40 + 0/2 x 40
therefore, s = 800m
Again,
v = u - gt
or, v = 40 - 10 x 40
therefore, v = -260 m/sec ( downward)
Total time taken to come back (T) = 40 s
Time taken to reach max. height (t) = 20 s
Consider distance b/w the ball thrown to max. height reached .
As the ball will attain rest at max. height , so
Final velocity (v) = 0 m/s
Time taken (t) = 20 s
As the ball is thrown against the gravity , so
Acceleration due to gravity (a) = -10 m/s²
⧪ As it is falling under and against the gravity , so we need to apply equations of kinematics ,
★ Apply 1st equation of motion ,
➠ v = u + at
➠ 0 = u + (-10)(20)
➠ 0 = u - 200
➠ u = 200 m/s
★ Apply 2nd equation of motion ,
➳ s = ut + ¹/₂ at²
➳ s = (200)(20) + ¹/₂ (-10)(20)²
➳ s = 4000 - 5(20)²
➳ s = 4000 - 5(400)
➳ s = 4000 - 2000
➳ s = 2000 m
So , Maximum height attained = 2000 m
Now , we need to find the velocity at which it hits the ground ,
Consider distance b/w max. height and bottom ,
As it starts from rest ,
Initial velocity (u) = 0 m/s
Time (t) = 20 s
As it is under gravity , so
Acceleration due to gravity , a = 10 m/s²
★ Apply 1st equation of motion ,
⭆ v = u + at
⭆ v = 0 + (10)(20)
⭆ v = 200 m/s
Velocity at which stone hits the ground is 200 m/s