Question

plzzzzz help me out... ​

Answer 1

Given:-

Initial velocity (u) = 50m/s

To Find:-

The maximum height reached by the body.

Time taken to reach the maximum height.

Solution:-

Initial velocity (u) = 50m/s

At maximum height velocity (v) = 0m/s

Acceleration (a) = g = -10m/s²

Let the maximum height be s

According to the third equation of motion:-

$\rm \dashrightarrow {v}^{2} = {u}^{2} + 2as$

Substituting the values:-

→ (0)² = (50)² + 2 × (-10) × s

→ 0 = 2500 + (-20)s

→ 0 = 2500 - 20s

→ 20s = 2500

→ s = $\dfrac{2500}{20}$

→ s = 125

$\underline{\boxed{\pink{\rm \therefore Maximum \: height = 125m}}}$

Now, let us find time:-

By first equation of motion:-

→ v = u + at

→ 0 = 50 + (-10)t

→ 0 = 50 - 10t

→ 10t = 50

→ t = $\dfrac{50}{10}$

→ t = 5

$\underline{\boxed{\purple{\rm \therefore Time \: taken \: to \: reach \: maximum \: height = 5sec}}}$

Answer 2

Answer:

Given:-

Initial velocity (u) = 50m/s

To Find:-

The maximum height reached by the body.

Time taken to reach the maximum height.

Solution:-

Initial velocity (u) = 50m/s

At maximum height velocity (v) = 0m/s

Acceleration (a) = g = -10m/s²

Let the maximum height be s

According to the third equation of motion:-

\rm \dashrightarrow {v}^{2} = {u}^{2} + 2as⇢v

2

=u

2

+2as

Substituting the values:-

→ (0)² = (50)² + 2 × (-10) × s

→ 0 = 2500 + (-20)s

→ 0 = 2500 - 20s

→ 20s = 2500

→ s = \dfrac{2500}{20}

20

2500

→ s = 125

$\underline{\boxed{\pink{\rm \therefore Maximum \: height = 125m}}}$

∴Maximumheight=125m

Now, let us find time:-

By first equation of motion:-

→ v = u + at

→ 0 = 50 + (-10)t

→ 0 = 50 - 10t

→ 10t = 50

→ t = \dfrac{50}{10}

10

50

→ t = 5

${\boxed{\purple{\rm \therefore Time \: taken \: to \: reach \: maximum \: height = 5sec}}}$

∴Timetakentoreachmaximumheight=5sec