A stone is thrown vertically upwards from top of a tower of 85m high, reaches the ground in 5s. Find
1. The greatest height above the ground.
2. The velocity with which it reaches the ground.
3. The time taken to reach the maximum height.
(g=10m/s)
A stone is thrown vertically upwards from top of a tower of 85m high, reaches the ground in 5s. Find
1. The greatest height above the ground.
2. The velocity with which it reaches the ground.
3. The time taken to reach the maximum height.
(g=10m/s)
The stone is thrown upward from height, h = 85 m from the ground.
It reaches the ground in 5 s.
Let the initial velocity be ‘u’.
The displacement is ‘-85 m’ and acceleration is ‘-10 m/s2’ as they are downward so the negative sign comes.
Using,
S = ut + ½ at2
=> -85 = 5u – ½ × 10 × 52
=> u = 8 m/s
At the greatest height above the ground, the stone comes to momentary rest. That means its velocity becomes zero.
Using,
v2 = u2 + 2as
=> 0 = 82 – 2 × 10 × s
=> s = 3.2 m
Thus, the maximum height from the ground is = 3.2 + 85 = 88.2 m
Velocity with which it reaches the ground is,
v = u + at
=> v = 8 – 10 × 5
=> v = -42 m/s [downward]
The stone is thrown upward from height, h = 85 m from the ground.
It reaches the ground in 5 s.
Let the initial velocity be ‘u’.
The displacement is ‘-85 m’ and acceleration is ‘-10 m/s2’ as they are downward so the negative sign comes.
Using,
S = ut + ½ at2
=> -85 = 5u – ½ × 10 × 52
=> u = 8 m/s
At the greatest height above the ground, the stone comes to momentary rest. That means its velocity becomes zero.
Using,
v2 = u2 + 2as
=> 0 = 82 – 2 × 10 × s
=> s = 3.2 m
Thus, the maximum height from the ground is = 3.2 + 85 = 88.2 m
Velocity with which it reaches the ground is,
v = u + at
=> v = 8 – 10 × 5
=> v = -42 m/s [downward]
