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2nd Equation of Motion

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The position of a projectile launched from the origin at t=0 is given byr=(40i + 50j)m at t=2s. If theprojectile was launched at an angle θ fromthe horizontal, then θ is (take g= 10 ms-2)(1) tan-1 2/3 (2)t an-1 3/2 (3) tan-17/4 (4) tan-1 4/5

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Q. The position of a projectile launched from the origin at

t = 0

\to r = (40^i + 50^j) m

t = 2 s

. If the projectile was launched at an angle

θ

from the horizontal, then

θ

(Take g = 10 {ms}^{- 2})

Q. The position of a projectile launched from the origin at

t = 0

\to r = (40^i + 50^j) m

t = 2 s

. If the projectile was launched at an angle

θ

from the horizontal, then

θ

g = 10 m s^{- 2})

.

Q. The position of a projectile launched from the origin at

t = o

\to r

(40^i + 50^i)

t = 2 s

. If the projectile was launched at an angle

θ

g = 10 m s^{- 2}) .

Q. The vertical height of the projectile at time

t

y = 4 t - 5 t^{2}

and the horizontal distance covered is given by

x = 3 t

x

y

m

t

s

. What is the angle of projection with the horizontal?

Q. A particle is projected with a velocity of 30 m/s, at an angle of

θ_{0} = t a n^{- 1} (3 / 4)

. After 1 second, the particle is moving at an angle

θ

to the horizontal, where tan

θ

(g = 10 m s^{- 2})

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