Question

A ball is thrown vertically up with a velocity of 49m/s.Calculate the maximum height to which at rises
बारिश टोन बचकानी अप विद ए वेलोसिटी ऑफ 49 मीटर पर सेकंड का लिटिल मैक्सिमम हाउ टू मेक ए टीचर ​

Answer 1

Given:-

• Initial velocity ,u = 49m/s

• Final velocity ,v = 0m/s

To Find:-

• Maximum height ,h

Solution:-

As we know that the acceleration due to gravity is 9.8 m/s²

By using 3rd equation of motion

• v² = u² + 2gh

Substitute the value we get

→ 0² = 49² + 2×(-9.8 )× h

→ 0 = 2401 + (-19.6 )×h

→ - 2401 = -19.6×h

→ h = -2401/-19.6

→ h = 2401/19.6

→ h = 122 .45 m

Therefore ,the maximum height attained by the ball is 122.45 metres.

Answer 2

Solution

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Given,

• the initial velocity = 49m/s
• the final velocity = 0 m/s

To find ,

• we have to find here the maximum height the ball reach

We know that ,

• acceleration due to gravity = 9.8m/s^2

Now,

• by using the third equation of motion we can find the value .

• v^2 = u^2 + 2gh

putting the value we get ;

$= > {0 }^{2} = {49}^{2} + 2 \times ( - 9.8) \times h$

$= > 0 = 2401 + ( - 19.6) \times h$

$= > 2401 = 19.6 \times h$

$= > h = \frac{ - 2401}{ - 19.6}$

$= > h = \frac{2401}{19.6}$

$= > h = 122.45m$

The ball will reach to 132•45m maximum height .

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