Question

The equations of motion of a projectile are given by $x=36t$m and $2y=96t-9.8t^2$ m. The angle of projection is

• A. sin${}^{-1}$(4/5)
• B. sin${}^{-1}$(3/5)
• C. sin${}^{-1}$(4/3)
• D. sin${}^{-1}$(3/4)

Answer 1

Answer: (A)

sin${}^{-1}$(4/5)

Given: $x-36t,2y=96t-9.8t^2$

or $y=48t-4.9t^2$

Let the initial velocity of projectile be u and angle of projection is $\theta$. Then,

Initial horizontal component of velocity,

$u_x=ucos\theta ={\frac{dx}{dt}}_{t=0}=36orucos\theta =36$---(i)

Initial vertical component of velocity,

$u_y=usin\theta ={\frac{dy}{dt}}_{t=0}=48orusin\theta =48$---(ii)

Dividing (ii) by (i), we get

$tan\theta =\frac{48}{36}=4/3$

$\therefore sin\theta =4/5$

or $\theta$=$si{n}^{-1}\left(4/5\right)$