a ball is thrown in upward direction from the tower with speed 10m/s and reaches the ground in 5s. find height of tower
Answers
Answer
- Height of the tower = 75 m
Given
- Initial velocity , u = 10 m/s
- Total time , t = 5 s
To Find
- Height of the tower , h
Concept Used
We need to apply equations of motion .
- v = u + at
- s = ut + ¹/₂ at²
- v² - u² = 2as
Solution
Height of the building , h = BC - AB = s" - s'
AB = BC [ Same height from horizontal ]
Find attachment for diagram of the situation .
Case-1 : AB
Initial velocity , u = 10 m/s
Final velocity , v = 0 m/s [ Goes to rest finally ]
Acceleration , a = g = - 10 m/s² [ Against the gravity ]
Distance AB , s' = ? m
Apply 3 rd equation of motion .
⇒ v² - u² = 2as
⇒ 0² - 10² = 2 ( - 10 ) s'
⇒ - 100 = - 20 s
⇒ s' = 5 m
Now , Time , t' = ? s
Apply 1st equation of motion .
⇒ v = u + at
⇒ 0 = 10 + ( - 10 )t'
⇒ 10t' = 10
⇒ t' = 1 s
______________________
Case-2 : BC
Initial velocity , u = 0 m/s [ Start's from rest ]
Time , t" = t - t' = 5 - 1 = 4 s
Acceleration , g = 10 m/s²
Distance , s" = ? m
Apply 2 nd equation of motion .
⇒ s = ut + ¹/₂ at²
⇒ s" = (0)t" + ¹/₂ (10)(t")²
⇒ s" = 5(4)²
⇒ s" = 5(16)
⇒ s" = 80 m
_______________________
So , Height of the building , h = s" - s'
⇒ h = 80 - 5
⇒ h = 75 m
Answer:
5m
Explanation:
here we are taking the first journey which means going up,
v(final velocity) = 0
u(initial velocity) = 10m/s
acceleration due to gravity = 10m/s (since it is going up)
using the equation v²-u²=2as
s = (v²-u²)÷2a
s = {(0)²-(10)²}÷2×(-10)
s = {-100}÷{-20}
s = 5m