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A ball is thrown with speed 20 m/s at an angle 60° with horizontal. After how much time will it move perpendicular to initial direction?

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4 years ago

Vikas TU
14146 Points
```							initial velocity makes 60 degree angle with horizontal.Thus at time t let the velocity is if makes 90 degree angle with initial velocity then the final velocity should make an angle 30 degree  with vertical.Hence from netns first law of eqn. we get,In x direcn. Vx = ucos60 = > u/2 => 20/2 = 10 m/sVy = -usin60 + gtort = root(3)/2g (Vy + u)...........................(1)Now we need to get Vy value.Now we know Vx => 10 = vsin30thus v = 20 m/sVy = vcos30 = > 20*cos30 => 10root(3) m/sPut Vy in eqn. (1) we get,t = (root(3)/20) (10root(3) + 20) secondsthat is approx. t = 3.226 seconds.
```
4 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