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Grade: 11
        
 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 
3 months ago

Answers : (2)

APARNA SHAJI
29 Points
							the answer is 600m/s approx . you have to consider both the vertical and horizontal velocities i.e, f
						
3 months ago
Khimraj
3008 Points
							

Horizontal Component

The horizontal component of the velocity is constant as it is perpendicular to g.

So we can write:

vcos30=20t

Where t is the time of flight.

v×0.866=20t

t=200.866v (1)

Vertical Component

We can use:

s=ut+12at2

This becomes:

5=vsin30t12gt2

5=v×0.5×t12×9.8×t2

5=v×0.5×t4.9t2 (2)

We can substitute the value of t from (1) into the first part of (2)

5=v×0.5×200.866v4.9t2

5=100.8664.9t2

4.9t2=11.5475=6.547

4.9t2=6.547

t2=6.5474.9=1.336

t=1.336=1.156xs

We 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

3 months ago
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