a ball is realeased from the top of a tower of height h.it takes T s to reach the ground. what is the position of the ball in T/3 s?
Answers
As the ball moves vertically, this is a case of free fall(motion under gravity in 1D).
For one dimensional motion we need a reference frame containing single axis and an origin.
Considering vertically downward direction as positive and the point of dropping the ball as the origin of this vertical axis.
The acceleration is constant throughout the motion so we can use equation of motion with constant acceleration.
Calculations
S=ut+1/2at^2
Here u =initial velocity =0
a= acceleration=g
t=time of motion
S=h = displacement along vertical axis = height of tower.
h=1/2gt^2 ………(1)
When the particle has moved for a time t/3 s.
Let the displacement is x
x= 1/2 g (t/3)^=(1/2gt^2)/9 ………(2)
From equation (1) putting value in (2)
Result
x=h/9
Position of the ball at time t/3 is at a depth h/9 from the top of the tower or at a height 8h/9 from the bottom of the tower.
Answer:
Explanation:
he acceleration of the ball will be g.
Initial velocity will be 0.
In T sec the body travels h.
by applying equations of motion we get
s= ut +1/2gT2
h = 1/2gT2 ------[1]
in T/3 sec h1 = 1/2gT2/9 -------[2]
from [1] and [2] we get h1 =h/9 distance from point of release.
therefore distance from ground is h-h/9 =8h/9