Question

A projectile is projected at an
angle 90° with the horizontal
and with a velocity of 3ms-1.
The velocity of the projectile at
the highest point of its
trajectory is
Select one:
O a.O
O b. 20ms 1
c. 15ms 1
O d. 30ms-1​

Answer 1

Answer:

Here u=30ms

−1

, Angle of projection, θ=90−30=60

∘

Maximum height,

H=

2g

u

2

sin

2

θ

=

2×10

30

2

sin

2

60

∘

=

20

900

×

4

3

=

4

135

m

Time of flight, T=

g

2usinθ

=

10

2×30×sin60

∘

=3

3

s

Horizontal range = R=

g

u

2

sin2θ

=

10

30×30×2sin60

∘

cos60

∘

=45

3

m

Explanation:

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Answer 2

Answer:

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