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`         Ball is thrown upward  at an angle of 30° with the horizontal and lands on the top edge of a building that is 20 m away the top touches 5 m above the throwing point the national speed of the ball in metre per second is  `
one year ago

APARNA SHAJI
32 Points
```							here we have two velocities one in the horizontal and othrt in the vertical direction so for the velocity of the particle you have to take the vector sum of it. H=U^2SIN^2 30/2g --(1) (vertiical)and R = U^2 SIN2*30 /g--(2)(horizontal).so you get U^2  ALONG VERTICAL DIRECTION AS 400 AND U^2 ALONG HORIZONTAL DIRECTION AS 400/ROOT3. SO BY GETTING THE ROOT MEAN SQUARE OF BOTH U GET THE ANSWER AS APPROX 600m/s
```
one year ago
Khimraj
3008 Points
```							Horizontal ComponentThe horizontal component of the velocity is constant as it is perpendicular to g.So we can write:vcos30=20tWhere t is the time of flight.∴v×0.866=20tt=200.866v (1)Vertical ComponentWe can use:s=ut+12at2This becomes:5=vsin30t−12gt2∴5=v×0.5×t−12×9.8×t25=v×0.5×t−4.9t2 (2)We can substitute the value of t from (1) into the first part of (2)⇒∴5=v×0.5×200.866v−4.9t2∴5=100.866−4.9t24.9t2=11.547−5=6.5474.9t2=6.547t2=6.5474.9=1.336t=√1.336=1.156xsWe can now put this value of t back into (1)⇒200.866v=t∴0.866v=20t=201.156∴v=201.156×0.866=19.97xm/s
```
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