A particle is thrown vertically up with an initial velocity 9 m/s form...
U= 9 m/s. g=10m/s
by applying 2nd eqn of motion
s=ut +1/2 at^2
we get
s= ut - 1/2gt^2 as body is subjected upward ,it's going opposite to gravity.
we get,
5t^2-9t+4=0
by solving we get
t=1 or t=0.1 s
A particle is thrown vertically up with an initial velocity 9 m/s form...
Given data:
Initial velocity (u) = 9 m/s
Acceleration due to gravity (g) = 10 m/s²
Height reached (h) = 4 m
Calculating time taken to reach the maximum height:
When the particle is thrown vertically up, it will reach the maximum height before falling down. At the maximum height, the particle's final velocity (v) will be zero.
Using the first equation of motion:
v = u + gt
At the maximum height, v = 0:
0 = 9 - 10t
10t = 9
t = 9/10
t = 0.9 s
So, it takes 0.9 seconds for the particle to reach the maximum height.
Calculating time taken to reach the height of 4 m for the second time:
To find the time taken to reach the height of 4 m again, we need to consider the total time taken for the particle to travel upwards and downwards.
1. Time taken to reach the maximum height (t₁) = 0.9 s (as calculated above)
2. Time taken to fall back from the maximum height (t₂) can be found using the second equation of motion:
h = ut + (1/2)gt²
4 = 0 + (1/2)(-10)t₂²
4 = -5t₂²
t₂² = -4/5
t₂ = √(-4/5)
Since time cannot be negative, t₂ will be imaginary. Therefore, the particle will not reach the height of 4 m again.
Summary:
- The time taken for the particle to reach the maximum height is 0.9 seconds.
- However, the particle will not reach the height of 4 m again as the time taken to fall back from the maximum height is imaginary.
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