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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`
4 years ago

Piyush
112 Points
```							Simply find out x when your y=2m Keeping time same for both the cases i.e when y1=2m and y2=2m and find out x1.
```
4 years ago
19 Points
```							Use equation of trajectory to find out the distance of the first hurdle from starting point. Y=xtan0-[g/u^2cos^2{theta} Put all the information givenU=10 m/sTheta-45g=10Y=2mWe get the answer-[x=2.75m]Then to find the distance between the two hurdles-D+2x=RWhere D=distance between the two hurdlesX=distance of first and last hurdle from starting and ending point respectivelyR=range of projectileUsing this we get d=4.5m
```
2 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