Question

a ball is projected from gorund st an angle theta with the horizontal. after 1s the ball is moving at an angle 45 with horizontal and after 3s it is loving horizontally. the velocity of protection of ball is​

Answer 1

Answer:

u=10√5m/s.

answered by anime lover.

Answer 2

Answer:

x and y components are as follows;-

x=ucosθt and vx=ucosθ

y=usinθt−

2

1

gt

2

⟶1

v

y

=usinθ−gt⟶2

The horizontal component of speed v

x

is always constant. So the speed is minimum when the vertical component v

y

=0

Putting this in equation 2

usinθ=2g

The ball is moving at angle 45° direction of velocity of the project at time t=1s

Then tan45°=1=

v

x

v

y

=

ucosθ

usinθ−g×1

⇒ucosθ=g

tanθ=2

θ=tan

−1

2

sinθ=

5

2

cosθ=

5

1

and u=

5

g

=

5

×10 m/s

u=10

5

m/s

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