Question

. An object is thrown up with a velocity of 98 m/s. Find the maximum

   height it reaches and also find the total time taken to return to the  

   starting point.  ​

Answer 1

$\huge {\underline{\pink{Given:-}}}$

• Initial velocity ,u = 98m/s
• Final Velocity ,v = 0m/s
• Acceleration due to gravity ,a = 9.8m/s²

$\huge {\underline{\green{To Find:-}}}$

• Maximum height attained by the object ,h
• Total time taken by the object.

$\huge {\underline{\blue{Solution:-}}}$

Firstly we calculate the height attained by the object.

Using 3rd Equation of Motion

• v² = u² +2ah

Substitute the value we get

➨ 0² = 98² + 2×(-9.8) ×h

➨ 0 = 9604 - 19.6 ×h

➨ -9604 = -19.6×h

➨ h = -9604/-19.6

➨ h = 9604/19.6

➨ h = 490m

Therefore , maximum height attained by the object is 490 Metres.

Now, Calculating the time taken by the object.

Using 1st Equation of motion

• v = u +at

Substitute the value we get

➨ 0 = 98 + (-9.8) ×t

➨ -98 = -9.8 ×t

➨ t = -98/-9.8

➨ t = 98/9.8

➨ t = 10 s

Therefore , the time taken by the object to reach maximum height is 10 seconds and again return its starting point so the total time taken by the object is 10 + 10 = 20 second .