Question

a projectile is thrown with velocity u=ai^ +bj^from ground of acceleration due to gravity is g then change in velocity of projectile in t second of projection is

Answer 1

\begin{lgathered}\vec{u}=a\ \vec{i}+b\ \vec{j}\\\\\frac{H}{R}=\frac{1}{4}Tan\theta=1/2\\\\so\ Tan\theta=2,\\\\from\ \vec{u},\ Tan\theta=\frac{b}{a}\\\\Hence, \ b=2a\end{lgathered}

u

=a

i

+b

j

R

H

=

4

1

Tanθ=1/2

so Tanθ=2,

from

u

, Tanθ=

a

b

Hence, b=2a

Answer 2

Explanation:

\begin{lgathered]\vec[u]-a\ \vec[i] +b\ \vec{j}\\\\\frac{H}{R}=\frac{1}{4} Tan\theta=1/2\\\\sol Tan\theta-2,\\\\from\ \vec{u},\ Tan\theta=\frac{b}{a}\\\\Hence, \ b=2a\end{lgathered]

Fa

i

+b

R

H

4

1

Tane=1/2

so Tane=2,

from

, Tane=

a

b

all 80% 1

Hence, b=2a