Question

An object thrown vertically upward reaches a height of 320m. What was it's initial velocity? How long will the object take to come back to the earth? Assume g=10m/s^2​

Answer 1

▩ Given :-

• Height Reached (s) = 320m
• Acceleration (g) = 10m/s²
• Final Velocity (v) = 0m/s

▩ To Find :-

• Time take by object to come back to the earth

▩ Solution :-

First, we need to find initial velocity (u),

By using third equation of motion,

$\sf \fbox{2as = {v}^{2} - {u}^{2}}$

$\sf →2 \times 10 \times 320 = {0}^{2} - {u}^{2}$

$\sf →6400 = 0 - {u}^{2}$

$\sf →6400 = - {u}^{2}$

$\sf → \sqrt{6400} = - u$

$\sf →\fbox \green{u = - 80m/s}$

∴ Initial velocity (u) is - 80m/s.

➸ Time

By using first equation of motion,

$\sf \fbox{v = u + at}$

$\sf →0 = - 80 + 10 \times t$

$\sf →0 + 80 = 10t$

$\sf →80 = 10t$

$\sf →t = \dfrac{8\cancel{0}}{1\cancel{0}}$

$\sf →\fbox \red{Time(t) = 8 \: seconds}$

∴The object will take 8 seconds to come back to the earth.

Answer 2

▩ Given :-

• Height Reached (s) = 320m
• Acceleration (g) = 10m/s²
• Final Velocity (v) = 0m/s

▩ To Find :-

Time take by object to come back to the earth

▩ Solution :-

First, we need to find initial velocity (u),

By using third equation of motion,

\sf \fbox{2as = {v}^{2} - {u}^{2}}

\sf →2 \times 10 \times 320 = {0}^{2} - {u}^{2}→2×10×320=0

2

−u

2

\sf →6400 = 0 - {u}^{2}→6400=0−u

2

\sf →6400 = - {u}^{2}→6400=−u

2

\sf → \sqrt{6400} = - u→

6400

=−u

\sf →\fbox \green{u = - 80m/s}→

u = - 80m/s

∴ Initial velocity (u) is - 80m/s.

➸ Time

By using first equation of motion,

v = u + at

\sf →0 = - 80 + 10 \times t→0=−80+10×t

\sf →0 + 80 = 10t→0+80=10t

\sf →80 = 10t→80=10t

\sf →t = \dfrac{8\cancel{0}}{1\cancel{0}}→t=

1

0

8

0

\sf →\fbox \red{Time(t) = 8 \: seconds}→

Time(t) = 8 seconds

∴The object will take 8 seconds to come back to the earth