Question

A projectile is thrown in the upward direction making an angle of 60$^{\circ}$ with the horizontal direction with a velocity of 147 $ms^{ - 1}$ Then the time after which its inclination with the horizontal is 45$^{\circ}$, is

• A. 15 s
• B. 10.98 s
• C. 5.49 s
• D. 2.745 s

Answer 1

Answer: (C)

Horizontal component of velocity at angle 60$^{\circ}$
= horizontal component of velocity at 45$^{\circ}$
ie, u cos 60$^{\circ}$ = v sin 45$^{\circ}$
or $ 147 \times \frac{ 1}{ 2} = v \times \frac{ 1}{ \sqrt 2 } $
or v = $ \frac{ 147}{ \sqrt 2 } \, ms^{ - 1} $
Vertical component of u = usin 60$^{\circ}$
$ \frac{ 147 \sqrt 3 }{ 2 } \, m $
Vertical component of v = vsin 45$^{\circ}$
$ = \frac{ 147 }{ \sqrt 2 } \times \frac {1}{ \sqrt 2} $
$ = \frac{ 147 }{ 2 } \, m $
but $ v_y = u_y + at $
$ \therefore \frac{ 147 }{ 2 } = \frac{ 147 \sqrt 3}{ 2 } - 9.8 t $
or 9.8 t = $ \frac{ 147 }{ 2 } ( \sqrt 3 - 1) $
$ \therefore t = 5.49 $ s