Question

a ball is projected vetically up with a speed of 50 m/s . find the maximum height , the time to reach the maximum height , and the speed at the maximum height​

Answer 1

Explanation:

v2-u2=2gh

0 - (50×50) = 2×10×h

h= 2500÷20=125m

v=u+at

0= 50-10t

10t=50

t= 5sec

Answer 2

Given :-

Initial velocity of the ball = 50 m/s

Final velocity at the highest point = 0 m/s

To Find :-

The maximum height.

The time to reach the maximum height.

The speed at the maximum height​.

Analysis :-

Here we are given with the initial velocity, final velocity and gravity.

Firstly using the third equation of motion substitute the given values in the question and find the maximum height.

Then using the first equation of motion find the time to reach the maximum height.

Finally, use the third equation of motion and find the speed at the maximum height​.

Solution :-

We know that,

• v = Final velocity
• g = Gravity
• h = Displacement
• u = Initial velocity

Using the formula,

$\underline{\boxed{\sf Third \ equation \ of \ motion=v^2=u^2-2gh}}$

Given that,

Final velocity (v) = 0 m/s

Initial velocity (u) = 50 m/s

Gravity (g) = 10 m/s

Substituting their values,

⇒ 0² = 50² - 2 × 10 × h

⇒ 0 = 2500 - 20 × h

⇒ h = 2500/20

⇒ h = 250/2

⇒ h = 125 m

Therefore, the maximum height is 125 m.

We know that,

• v = Initial velocity
• g = Gravity
• v = Final velocity
• t = Time

Using the formula,

$\underline{\boxed{\sf First \ equation \ of \ motion=v=u+gt}}$

Given that,

Final velocity (v) = 0 m/s

Initial velocity (u) = 50 m/s

Gravity (g) = 10 m/s

Substituting their values,

⇒ 0 = 50 + 10 × t

⇒ 10t + 50 = 0

⇒ 10t = 50

⇒ t = 50/10

⇒ t = 5 sec

Therefore, the time to reach the maximum height is 5 sec.

We know that,

• v = Final velocity
• g = Gravity
• h = Displacement
• u = Initial velocity

Using the formula,

$\underline{\boxed{\sf Third \ equation \ of \ motion=v^2=u^2-2gh}}$

Given that,

Initial velocity (u) = 50 m/s

Gravity (g) = 10 m/s

Displacement (s) = 125 m

Substituting their values,

⇒ v² = 50² - 2 × 10 × 125

⇒ v² = 2500 - 2500

⇒ v² = 0 m/s

Therefore, the speed at the maximum height​ is 0 m/s.