An object is thrown vertically upwards and rises to a height of 45 m ,take g=10m/s^2.Calculate total time for which object remains in air.
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An object is thrown vertically upwards and rises to a height of 45 m ,take g=10m/s^2.Calculate total time for which object remains in air.
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Given:
Maximum height attained by object, h= 45m
To Find:
Total time for which object remains in air
Solution:
We know that,
[tex]\purple{\boxed{\bf{v=u+at}}}[/tex]
[tex]\orange{\boxed{\bf{s=ut+\dfrac{1}{2}at^{2}}}}[/tex]
[tex]\pink{\boxed{\bf{v^{2}-u^{2}=2as}}}[/tex]
where,
v is final velocity
u is initial velocity
a is acceleration
s is displacement
[tex]\rule{190}{1}[/tex]
Reference taken here:
[tex]\rule{190}{1}[/tex]
We have to consider two cases:
Case-1: When object is going upward
Let the time taken by object to reach the highest point be t₁, initial velocity be u, final velocity be v and s be the displacement of object
On applying third equation of motion for upward motion of object, we get
[tex]\longrightarrow\rm{v^{2}-u^{2}=2as}[/tex]
[tex]\longrightarrow\rm{(0)^{2}-u^{2}=2(-g)(45)}[/tex]
[tex]\longrightarrow\rm{-u^{2}=-90g}[/tex]
[tex]\longrightarrow\rm{u^{2}=90(10)}[/tex]
[tex]\longrightarrow\rm{u^{2}=900}[/tex]
[tex]\longrightarrow\rm{u=\sqrt{900}}[/tex]
[tex]\longrightarrow\rm{u=30\:m/s}[/tex]
On applying first equation of motion for upward motion of object, we get
[tex]\longrightarrow\rm{v=u+at}[/tex]
[tex]\longrightarrow\rm{0=30+(-g)t_{1}}[/tex]
[tex]\longrightarrow\rm{0=30-gt_{1}}[/tex]
[tex]\longrightarrow\rm{gt_{1}=30}[/tex]
[tex]\longrightarrow\rm{10(t_{1})=30}[/tex]
[tex]\longrightarrow\rm{t_{1}=\dfrac{30}{10}}[/tex]
[tex]\longrightarrow\rm{t_{1}=3\:s}[/tex]
[tex]\rule{190}{1}[/tex]
Case-2: When object is going downward
Let the time taken by object to reach the ground be t₂, initial velocity be u' and s be the displacement of the object
On applying second equation of motion for downward motion of object, we get
[tex]\longrightarrow\rm{s=ut+\dfrac{1}{2}at^{2}}[/tex]
[tex]\longrightarrow\rm{-45=0(t)+\dfrac{1}{2}(-g)(t_{2})^{2}}[/tex]
[tex]\longrightarrow\rm{-45=\dfrac{1}{2}(-10)(t_{2})^{2}}[/tex]
[tex]\longrightarrow\rm{-45=-5(t_{2})^{2}}[/tex]
[tex]\longrightarrow\rm{-5(t_{2})^{2}=-45}[/tex]
[tex]\longrightarrow\rm{(t_{2})^{2}=\dfrac{\cancel{-45}}{\cancel{-5}}}[/tex]
[tex]\longrightarrow\rm{(t_{2})^{2}=9}[/tex]
[tex]\longrightarrow\rm{t_{2}=\sqrt{9}}[/tex]
[tex]\longrightarrow\rm{t_{2}=3\:s}[/tex]
[tex]\rule{190}{1}[/tex]
So,
[tex]\rm{Total\:time\:taken=t_{1}+t_{2}}[/tex]
[tex]\rm{Total\:time\:taken=3+3}[/tex]
[tex]\rm\green{Total\:time\:taken=6\:s}[/tex]
Hence, total time for which object remains in air is 6 s.
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Here's ur answer
Displacement = S = h = 45m
Acceleration = a = g = 10m/s²
Initial velocity = u = 0 m/s
Time taken for the body to reach the highest point from the bottom is equal to the time taken for it to come down
We know that,
S = ut + 1/2 at²
45 = 0 + 1/2 * 10 * t²
t² = 45 * 2/10 = 90/10 = 9
t = 3s
Therefore the body takes 3s to reach highest point and 3s to come down
Hence,
Total time that it stays in the air = 3+3 = 6s
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