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`         Abolish thrown up words at an angle of 30° with the horizontal and lands on the top of a building that is 20 m away the top edges 5 m above the throwing point initial speed of the ball in metre per second is `
one year ago

APARNA SHAJI
32 Points
```							the answer is 600m/s approx . you have to consider both the vertical and horizontal velocities i.e, f
```
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