Question

A projectile is thrown with velocity v at an angle theta with the horizontal. When the projectile is at a height equal to half of the maximum height,.
The vertical component of the velocity of projectile is.

Answer 1

Answer:

A projectile is thrown with velocity v at an angle θ with horizontal.

then, v

x

=vcosθ,v

y

=vcosθ

we have to find vertical component of velocity of the projectile at height equal to half of the maximum height.

we know, H

max

=

2g

u

2

sin

2

θ

here u is intial velocity of projectile. but here given initial velocity is v

so, H

max

=

2g

v

2

sin

2

θ

now, time taken to reach half of maximum height, t

Y=v

y

t+

2

1

a

y

t

2

2g

v

2

sin

2

θ

=vsinθ.t−5t

2

200t

2

−40vsinθ.t+v

2

sin

2

θ=0

t=

10

2

(

2

±1)vsinθ

v

y

′

=v

y

+a

y

t

=vsinθ−

2

(

2

−1)vsinθ

=

2

vsinθ

Answer 2

Answer:

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