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`        Que. A stone is thrown vertically upward with velocity of 20 m per second from 25 m top of Tower. Determine - (1) the distance covered by stone before reaching the ground. (2) maximum height attained during flight. (3) final velocity of stone before reaching the ground.`
2 years ago

53 Points
```							the stone is thrown with initial velocity 20 m/sit is thrown under gravity so gravitational accleration will be9.8 m/s and final velocity will be 0therefore bu using v2=u2+2as where v=final velocity, u=initial velocity, a= gravitational acceleration, s= distance covered.after putting values in the eqn and solving we get distance covered=s=20.4 mso the stone has covered a dist of 20.4m after throwing in the vertical directionso its new height is 25+20.4=45.4m this the max heightbefore reaching the ground it will travel a distance of 20.4+20.4+25=65.8mnow for its downward journey, initial velocity=u=0final velocity=vacceleration=9.8 dist to be covered = 45.4mtherefore putting above values in the eqn     v2=u2 +2ason solving we get   v2=889.84so v=29.83
```
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