Question

A ball is projected from ground at an angle 0 with the
horizontal. After 1 s, ball is moving at an angle 45° with
the horizontal and after 3 s it is moving horizontally.
The velocity of projection of ball is (g = 10 m/s?)​

Answer 1

Answer:

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Explanation:

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