Aisha dropped an object of mass 4 Kg from a height of 10m.
The object has a speed of 0 m/s at position A and is at a height
of 10m above the ground. At position B, the object is 6m above
the ground. At position C, the object is 4m above the ground.
6
At point D, the object is mere picometer ( almost at 0 meter height)
above the ground. Assume negligible air resistance throughout
the motion. Use this information to fill in the table. (g=10m/s2)
Answers
Given : The object has a speed of 0 m/s at position A and is at a height
A of 10m above the ground.
To find : Velocity & time taken to reach at B , C & D
Solution:
At t = 0 position A - 10 m above ground velocity = 0
PE = mgh = 4 * 10 * 10 = 400 Joule
KE =(1/2)mv² = 0
At position B - 6 m above ground distance covered (10 - 6) = 4 m
S = ut + (1/2)at²
u = 0
a = g = 10
=> 4 = (1/2)(10)t²
=> t = √0.8 = 0.9 sec
V = U + at = 9 m/s
PE = 4 * 10 * 6 ≈ 240 joule
KE = 400 - 240 ≈ 160 Joule
At position C - 4 m above ground distance covered (10 - 4) =6 m
S = ut + (1/2)at²
u = 0
a = g = 10
=> 6 = (1/2)(10)t²
=> t = √1.2 ≈ 1.1 sec
V = U + at ≈ 11 m/s
PE = 4 * 10 * 4 = 160 joule
KE = 400 - 160 = 240 Joule
At position D - 0 m above ground distance covered (10 - 0) =10 m
S = ut + (1/2)at²
u = 0
a = g = 10
=> 10 = (1/2)(10)t²
=> t = √2 = 1.4sec
V = U + at = 14 m/s
PE = 4 * 10 * 0 = 0 joule
KE = 400 Joule
Position Distance from ground PE KE Time Velocity
A 10 m 400 0 0 s 0 m/s
B 6 m 240 160 0.9 s 9 m/s
C 4 m 160 240 1.1 s 11 m/s
D 0 m 0 400 1.4 s 14 m/s
Given :-
The object has a speed of 0 m/s at position A and is at a height
A of 10m above the ground.
To find :-
Velocity & time taken to reach at B , C & D
Solution :-
At t = 0 position A - 10 m above ground velocity = 0
PE = mgh = 4 * 10 * 10 = 400 Joule
KE =(1/2)mv² = 0
At position B - 6 m above ground distance covered (10 - 6) = 4 m
S = ut + (1/2)at²
u = 0
a = g = 10
=> 4 = (1/2)(10)t²
=> t = √0.8 = 0.9 sec
V = U + at = 9 m/s
PE = 4 * 10 * 6 ≈ 240 joule
KE = 400 - 240 ≈ 160 Joule
At position C - 4 m above ground distance covered (10 - 4) =6 m
S = ut + (1/2)at²
u = 0
a = g = 10
=> 6 = (1/2)(10)t²
=> t = √1.2 ≈ 1.1 sec
V = U + at ≈ 11 m/s
PE = 4 * 10 * 4 = 160 joule
KE = 400 - 160 = 240 Joule
At position D - 0 m above ground distance covered (10 - 0) =10 m
S = ut + (1/2)at²
u = 0
a = g = 10
=> 10 = (1/2)(10)t²
=> t = √2 = 1.4sec
V = U + at = 14 m/s
PE = 4 * 10 * 0 = 0 joule
KE = 400 Joule
Position Distance from ground PE KE Time Velocity
A 10 m 400 0 0 s 0 m/s
B 6 m 240 160 0.9 s 9 m/s
C 4 m 160 240 1.1 s 11 m/s
D 0 m 0 400 1.4 s 14 m/s
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