Slide 1 : P. K. BHARTI, I.I.T. KHARAGPUR KINEMATICS
Acceleration due to Gravity
(g) © 2007-10 P K Bharti, IIT Kharagpur
Slide 2 : © 2007-10 P K Bharti, IIT Kharagpur
Slide 3 : Q. A ball is dropped from a balloon going up at a speed of 7 m/s. If the balloon was at a height 60 m at the time of dropping the ball, how long will the ball take reaching the ground?
SOLUTION:
Let us see the animation.
The ball starts moving upward with initial velocity 7 m/s because of inertia, goes to maximum height and then starts coming down.
Net time taken by ball to reach the ground
= (time taken by ball to go to cover AB)
+ (time taken by ball to go to cover BC).
Part AB:
Initial velocity: u1 = 7m/s (upward)
Final velocity: v1 = 0
Acceleration: a1 = – g = – 9.8 m/s2 (downward)
Let the time taken by ball to cover AB be t1.
Then using, v = u + at , we get
v1 = u1 + a1t1 ? 0 = 7 – 9.8t1 ? t1 = 0.71s Ground A C B
Slide 4 : Again using s = ut + ½ at2 , we get
AB = u1t1 + ½ a1 t1 2
? AB = 7 × 0.71 + ½ × (– 9.8) ×(.71) 2 = 2.5 m
Part BC:
Initial velocity u2 = 0
Total distance BC = AB + AC = 62.5 m
Acceleration a2 = g = 9.8 m/s2
Using s = ut + ½ at2
? BC = u2t2 + ½ a2 t2 2
? 62.5 = 0 × t2 + ½ × (– 9.8) × t22
? t22 = 12.755
? t2 = ± 3.57.
Neglecting negative value t2 = 3.57 s.
Net time taken by ball to reach the ground
= t1 + t2 = 0.71 + 3. 57 = 4.28 s
˜ 4.3s (Ans) Ground C B A
Slide 5 : Q. An elevator is descending with uniform acceleration. To measure the acceleration, a person in the elevator drop a coin at the moment the elevator starts. The coin is 6ft above the floor of the elevator at the time it is dropped. The person observes that the coin strikes the ground in 1s. Calculate from these data the acceleration of the elevator.
SOLUTION:
Let us visualize the situation.
ELEVATOR:
Let the acceleration of the lift be a ft/s2.
Let the elevator moves a distance x ft downward.
Now, for elevator we have,
distance = x ft (downward)
acceleration = a ft/s2
initial velocity = 0
time = 1 s
Thus using, s = ut + ½ at2 , we have,
x = 0 × 1 + ½ × a × 12
? x = ½ a …(1) x ft 6 ft
Slide 6 : COIN:
The coin fall freely under the action of gravity.
Therefore, its acceleration = 32 ft/s2 (downward).
NOTE: g = 9.8 m/s2 = 32 ft/s2
As the elevator moves a distance x ft downward, the coin moves (x+6) ft downward.
Now, for coin we have,
distance = (x+6) ft (downward)
acceleration = 32 ft/s2
initial velocity = 0
time = 1 s
Thus using, s = ut + ½ at2 , we have,
x + 6 = 0 × 1 + ½ × 32 × 12
? x = 10 …(2)
Slide 7 : Putting this value of x = 10 in (1) we get ,
10 = ½ a
? a = 20
Therefore, acceleration of elevator = 20 ft/s2. (Ans)
Slide 8 : Q. A block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration of 12 m/s2. Find the displacement of the block during first 0.2 s after the start. Take g = 10 m/s2.
20 cm
24 cm
4 cm
44 cm