A ball is dropped from a tower .in the last second of it motion it travels a distances of 15m . find the hight of tower .
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Answered by
302
Heya user !!
Here's the answer you are looking for
Let the height of the tower is h and time taken by it to reach the ground from height h is t.
Since, it is dropped, it's initial velocity (u) = 0
Using the second equation of motion,
The last second means at t = √(h/5)th second.
So, at t = √(h/5)th second, the ball travels 15m
Therefore, the height of the tower is 20m.
★★ HOPE THAT HELPS ☺️ ★★
Here's the answer you are looking for
Let the height of the tower is h and time taken by it to reach the ground from height h is t.
Since, it is dropped, it's initial velocity (u) = 0
Using the second equation of motion,
The last second means at t = √(h/5)th second.
So, at t = √(h/5)th second, the ball travels 15m
Therefore, the height of the tower is 20m.
★★ HOPE THAT HELPS ☺️ ★★
Anshukumar111:
Thanku
Answered by
2
Answer:
The height of tower is 20 m
Explanation:
- Now as we know that the equation of motion is
and for nth second we have the equation of motion as
- So by assuming height of the building to be h and total time taken by the ball to reach ground be t and initial velocity is 0 hence we get the equation as
;
on solving this we get - Now in last second that is th second it travelled 15 m so
- After solving this we get the value of h as 20m
#SPJ2
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