Question

2. A stone thrown upwards from the top of a tower 85 m high, reaches the ground in 5 s. Find (i) the greatest height above the ground ​

Answer 1

First draw the fig as per the ques.

Now according to attached image:

• AB is height of tower = 85m
• Its maximum height will be reach at topmost point 'C' where its velocity will be zero.
• So the greatest height above the ground will be AC.

Solution:

•Net displacement from A to E = -85 m(downward)

(as BC and CD get cancelled during the motion due to opposite direction)

• and whole motion fron A to E takes 5s

1. s = ut + ½ at²
2. -85 = u(5) + ½(-10)(5)²
3. -85 = 5u - 5(25)
4. -17 = u - 25
5. u = 8 m/s

•Let's calculate displacement in BC

1. velocity at B = u = 8 m/s
2. velocity at C = v = 0 m/s
3. v² = u² + 2as
4. 0² = (8)² + 2(-10)s
5. 0 = 64 - 20s
6. -64 = -20s
7. s= 64/20 = 3.2 m

》Maximum height = AB + BC

= 85 + 3.2

= 88.2 m

(taking g as 10m/s² , if you take 9.8 m/s² then answer will be slightly different)

Answer 2

Explanation:

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• First draw the fig as per the ques.
• Now according to attached image:
• AB is height of tower = 85m
• Its maximum height will be reach at topmost point 'C' where its velocity will be zero.
• So the greatest height above the ground will be AC.

Solution:

•Net displacement from A to E = -85 m(downward)

(as BC and CD get cancelled during the motion due to opposite direction)

• and whole motion fron A to E takes 5s

s = ut + ½ at²

-85 = u(5) + ½(-10)(5)²

-85 = 5u - 5(25)

-17 = u - 25

u = 8 m/s

•Let's calculate displacement in BC

velocity at B = u = 8 m/s

velocity at C = v = 0 m/s

v² = u² + 2as

0² = (8)² + 2(-10)s

0 = 64 - 20s

-64 = -20s

s= 64/20 = 3.2 m

》Maximum height = AB + BC

= 85 + 3.2

= 88.2 m

(taking g as 10m/s² , if you take 9.8 m/s² then answer will be slightly different)

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