Question

a ball is projected from the ground at angle theta with the horizontal after 1s it is. moving at angle 45degree with the horizontal and after 2s it is moving horizontally what is the velocity of projecrion of ball???????

Answer 1

speed is minimum after 2 sec means speed=vcos theta

after 1 second

v'cos45=vcos theta

also

v'sin45-g*1=0

=>v'sin45=g

=>v'=gsqrt2

also

vsin theta-g*1=v'sin45=g

=>vsin theta=2g

=>tan theta=2g/v'cos45=2

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