Question

An object is thrown is vertically upwards and rises to a height of 45 m . calculate [take g= 10m/s^2]
i) the velocity with which the object was thrown upward.
ii) the time taken by object to reach highest point.
iii) total time for object which remains in air.
plz answer it correctly

Answer 1

Answer:

i. v^2 - u^2 = 2gh

0^2 - u^2 = 2*-10m/s^2 * 45m

-u^2 = -900m^2/s^2

u = 30m/s

ii. t = v - u / g

t = 0 - 30m/s / -10m/s^2

t = 3s

iii. .......................

hope it helps

Answer 2

Given:

Maximum height attained by object, h= 45m

To Find:

i) The velocity with which the object was thrown upward.

ii) The time taken by object to reach highest point.

iii) 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 velocities, forces and accelerations acting in upward direction are taken positive.

• All 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\rule{190}{1}$

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

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

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

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

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

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

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

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

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

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

$\rule{190}{1}$

Le the total time for which object remains in air be t

So,

$\longrightarrow\rm{t=t_{1}+t_{2}}$

$\longrightarrow\rm{t=3+3}$

$\longrightarrow\rm\green{t=6\:s}$

Hence,

the velocity with which the object was thrown upward is 30 m/s

the time taken by object to reach highest point is 3s

the total time for which object remains in air is 6s