a particle thrown vertically upwards with a velocity of 14.7 meter per second from a point 19.6 above the ground find the time taken by it to reach the ground?given g=9.8meter per second
Answers
Explanation:
u = 14.7 m/s.
v = 0 m/s.
h = 19.6m.
g = 9.8 m/s^2
by using v = u + gt
0 = 14.7 + 9.8 × t
9.8 t = 14.7
t = 14.7/9.8
t = 1.5 s
So , it takes 1.5 second to reach the ground.
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Given:
The initial velocity of the particle, u = 14.7 m/s
Height above the ground, h = 19.6 m
Acceleration due to gravity, g = 9.8 m/s²
Find:
The time taken by it to reach the ground.
Solution:
Particle is thrown vertically upwards from the height of 19.6 m.
Height, h = 19.6 m
Case I: Particle moving upward. Let the maximum height attained by it be 's'. At maximum height, the particle will be at rest and then will fall freely.
Final velocity, v = 0 m/s
Initial velocity, u = 14.7 m/s
Acceleration, a = -g = -9.8 m/s²
Let the time taken by it to cover distance s be t₁.
According to first equation of motion, we have
v = u + at₁
0 = 14.7 + (-9.8)t₁
14.7 - 9.8t₁ = 0
9.8t₁ = 14.7
t₁ = 14.7/9.8
t₁ = 1.5 seconds
According to third equation of motion, we have
v² - u² = 2as
(0)² - (14.7)² = 2(-9.8)s
-216.09 = -19.6 s
s = 216.09/19.6
s = 11.025 m
Case II: After attaining maximum height, the particle falls freely under the force of gravity.
Initial velocity, u = 0 m/s
Acceleration = g = 9.8 m/s²
Total distance, S = s + h
= 11.025 + 19.6
= 30.625 m
Let the time taken by the particle to cover the distance of S = 30.625 m be t₂.
According to the second equation of motion, we have
S = ut₂ + 1/2at₂²
30.625 = (0)t + 1/2 (9.8)t₂²
30.625 = 0 + 4.9t₂²
4.9t₂² = 30.625
t₂² = 30.625/4.9
t₂² = 6.25
t₂ = √6.25
t₂ = 2.5 seconds
∴ The time taken by the particle to reach the ground, T = t₁ + t₂
= 1.5 + 2.5
= 4 seconds
Hence, the time taken by the particle to reach the ground is 4 seconds.
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