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A projectile is thrown at an angle of 60° to the horizontal,strikes a wall 30m away at a point 15m above the point of projection.....(i) find the speed of projection

```
3 years ago

Aman Sharma
40 Points
```							We know that the distance covered vertically by the projectile is 15m  so by equation of motion ( 2aysy = vy2 - uy2) we have s = 15m   a=g  u= usin60   thus (usin60)2 = 2g15m solving we get u2= 40g m    let the value of g be 10ms--2  thus u2=400ms--1  by square rooting on both sides we get u=20ms--1
```
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