Question

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.

Answer 1

Given:

Maximum height attained by object, h= 45m

To Find:

Total time for which object remains in air

Solution:

We know that,

• When a body is thrown vertically upwards, then velocity of body at highest point is 0
• According to first equation of motion for constant acceleration,

$\purple{\boxed{\bf{v=u+at}}}$

• According to second equation of motion for constant acceleration,

$\orange{\boxed{\bf{s=ut+\dfrac{1}{2}at^{2}}}}$

• According to third equation of motion for constant acceleration,

$\pink{\boxed{\bf{v^{2}-u^{2}=2as}}}$

where,

v is final velocity

u is initial velocity

a is acceleration

s is displacement

$\rule{190}{1}$

Reference taken here:

• All displacements, velocities, forces and accelerations acting in upward direction are taken positive.

• All displacements, velocities, forces and accelerations acting in downward direction are taken negative.

$\rule{190}{1}$

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

$\longrightarrow\rm{v^{2}-u^{2}=2as}$

$\longrightarrow\rm{(0)^{2}-u^{2}=2(-g)(45)}$

$\longrightarrow\rm{-u^{2}=-90g}$

$\longrightarrow\rm{u^{2}=90(10)}$

$\longrightarrow\rm{u^{2}=900}$

$\longrightarrow\rm{u=\sqrt{900}}$

$\longrightarrow\rm{u=30\:m/s}$

On applying first equation of motion for upward motion of object, we get

$\longrightarrow\rm{v=u+at}$

$\longrightarrow\rm{0=30+(-g)t_{1}}$

$\longrightarrow\rm{0=30-gt_{1}}$

$\longrightarrow\rm{gt_{1}=30}$

$\longrightarrow\rm{10(t_{1})=30}$

$\longrightarrow\rm{t_{1}=\dfrac{30}{10}}$

$\longrightarrow\rm{t_{1}=3\:s}$

$\rule{190}{1}$

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

$\longrightarrow\rm{s=ut+\dfrac{1}{2}at^{2}}$

$\longrightarrow\rm{-45=0(t)+\dfrac{1}{2}(-g)(t_{2})^{2}}$

$\longrightarrow\rm{-45=\dfrac{1}{2}(-10)(t_{2})^{2}}$

$\longrightarrow\rm{-45=-5(t_{2})^{2}}$

$\longrightarrow\rm{-5(t_{2})^{2}=-45}$

$\longrightarrow\rm{(t_{2})^{2}=\dfrac{\cancel{-45}}{\cancel{-5}}}$

$\longrightarrow\rm{(t_{2})^{2}=9}$

$\longrightarrow\rm{t_{2}=\sqrt{9}}$

$\longrightarrow\rm{t_{2}=3\:s}$

$\rule{190}{1}$

So,

$\rm{Total\:time\:taken=t_{1}+t_{2}}$

$\rm{Total\:time\:taken=3+3}$

$\rm\green{Total\:time\:taken=6\:s}$

Hence, total time for which object remains in air is 6 s.

Answer 2

Gud Evening

How r u?? ❤️

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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