A stone thrown upward from the top of the tower 75m height reached the ground in 6sec find the velocity with which it is thrown upwards
Answers
Answer:
The velocity with which the stone is thrown upwards is 44.6 m/s
Explanation:
Height of the tower, h = 75 m
The stone is thrown upward and it reached the ground in time, t = 6 s
The distance traveled by the stone from its maximum height to the ground can be calculated as:
S = ut + [(at²)/2]
Initial velocity is 0 as the stone is under free fall from its maximum height.
∴ S = 0 + [(9.8 × 6²)/2]
S = 176.4 m
From this value, the height the stone was thrown from the top of the tower can be calculated as:
H = S - h
= 176.4 - 75
H = 101.4 m
So 101.4 m is the distance traveled by the stone from the tower to its maximum height. The final velocity of the stone at maximum height will be zero. Now initial velocity can be found using the 3rd equation of motion.
v² - u² = 2aS
⇒ u² = v² - 2aS
u² = 0² - (2 × -9.8 × 101.4)
[Value of g is negative as the stone is going upwards against the force of gravity.]
u² = 1987.44
∴ u = √1987.44
= 44.6 m/s