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`        A projectile takes off with an initial velocity of 10m/s at an angle of elevation of 45° .it is just able to clear two hurdles of height 2 m each,separated from each other by a distance d.calculate d.at what distance from the point of projection is the first hurdle placed? Take g= 10 m/s^2`
one year ago

Arun
24739 Points
```							Use the trajectory equation: y = h + x·tanΘ - g·x² / (2v²·cos²Θ) where y = height at x-value of interest = 2 m and h = initial height = 0 m and x = range of interest = ??? and Θ = launch angle = 45º and v = launch velocity = 10 m/s Dropping units for ease, 2 = 0 + xtan45 - 9.8x²/(2(10)²cos²45) = x - 0.098x² This quadratic has roots at x = 2.73 m ← first hurdle and x = 7.47 m ← second hurdle d = 7.47m - 2.73m = 4.74 m ◄
```
one year 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