Question

a projectile is thrown at an angle theta with the horizontal with speed u after some time it makes angle alpha with horizontal the speed of the projectile at that instant is



​

Answer 1

Answer:

Horizontal component of velocity =v

x

=ucosθ

Vertical component of velocity v

y

=usinθ−gt

angle of velocity with horizontal =α

tana=

v

x

v

y

=

(4cosθ)

(usinθ−gt)

t=

g

u(sinθ−cosθtanα)

v

y

=ucosθtanα

v

x

=ucosθ

So the speed of the projectile =

((v

x

)

2

+(v

y

)

2

)

=ucosθsecα

Answer 2

Answer:

Its speed when its direction of motion makes an angle 'α' with the horizontal is.

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