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```
a ball is thrown so that it just clears a 3 meter fence 18 meters away. If it left the hand 1.5 meters above the ground and at an angle of 60 o with the horizontal, what was the initial velocity of the ball?
a ball is thrown so that it just clears a 3 meter fence 18 meters away. If it left the hand 1.5 meters above the ground and at an angle of 60o with the horizontal, what was the initial velocity of the ball?

```
3 years ago

Sounak Dutta
27 Points
```							Resolving into components we will find that horizontal velocity is v*cos600 whereas vertical component is v*sin600....(we are assuming the velocity of the ball as v initially)So...forming equations let the ball reach the wall in time t.     so along horizontal                     v*cos600*t=18             as distance between thrower and wall is 18 metres.    Along vertical,                 using formula v2=u2-2*a*s..      here u being v*sin600 and v being 0,,a being g,i.e., 9.8m/s        s = 1.5m we get another equation in v... but...by only solving second equation....we get our required velocity as 39.2 m/s
```
3 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions