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

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