Question

A body is projected with initial velocity of 40 m/s and at an angle of 30°with the horizontal.

Calculate (i) time of flight and (ii) horizontal range of projectile.(assume g= 10 m/s2

)​

Answer 1

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Answer 2

Explanation:

R.E.F image

u=40m/s

θ=30

∘

g=9.81

Time of flight, T=

g

2usinθ

Maximum Height, H=

2g

u

2

sin

2

θ

Horizontal Range, R=

g

u

2

sin2θ

Time taken to reach the

max.height, t =

2

T

4=

2×9.8

(40)

2

(sin30

∘

)

2

=20.408m

=20.41m

R=

9.8

(40)

2

sin2×30

∘

=141.4m

t=

2

I

=

g

usin3θ

=2.041s

Ans =141.4m,20.41m,2.041s (C)