Question

An object thrown vertically upwards
reaches a height of 500 m. What was
its initial velocity? How long will the
object take to come back to the
earth? Assume g = 10 m/s
​

Answer 1

GIVEN :

• An object thrown vertically upward reaches a height of 500 m.

TO FIND :

• Initial velocity = ?
• How long will the object take to come back on the earth ?

FORMULA :

• Newton's first law of motion, $\sf V \: = \: u \: + \: at$
• Newton's third law of motion, $\sf 2gs \: = \: v^{2} \: - \: u^{2}$

SOLUTION :

» Initial velocity, u

» Final velocity, v = o m/s.

» Distance, S = 500 m.

Using Newton's third law of motion,

$\red {\boxed {\sf 2gs \: = \: v^{2} \: - \: u^{2}}}$

Put the values in the above equation,

→ $\sf 2 \: (-10) \: 500 \: = \: u^{2}$

$\therefore \boxed{\sf u \: = \: 100 \: m/s}$

$\therefore$ Its initial velocity (u) was 100 m/s.

Now,

To find the time, t = ?

Using Newton's first law of motion,

$\red {\boxed {\sf V \: = \: u \: + \: at}}$

Put the values in the above equation,

→ $\sf 0 \: = \: 100 \: - \: 10 t$

→ $\sf t \: = \: \dfrac {100}{10}$

→ $\sf t \: = \: \dfrac 10 \: sec$

$\therefore \boxed{\sf t \: = \: 10 \: sec}$

•°• The initial velocity was 100 m/s and time taken was 10 + 10 = 20 sec.

Answer 2

Answer:

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