Math, asked by sanjayraj7907, 1 year ago

A ball is thrown vertically upward at an angle of 30 with the horizontal and lands on the top edge of a builing that is 20m away

Answers

Answered by Anonymous
2
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=vsin30t−12gt2

∴5=v×0.5×t−12×9.8×t2

5=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=100.866−4.9t2

4.9t2=11.547−5=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

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