Question

A projectile is thrown with a speed of $100m/s$ making an angle of ${60}^{\circ }$ with the horizontal. Find the time after which its inclination with the horizontal is ${45}^{\circ }$ ?

• A. $5(\sqrt{3}-1)$
• B. $5(\sqrt{\frac{3}{2}}-1)$
• C. $5\sqrt{3}$
• D. $5(\sqrt{3}-2)$

Answer 1

The correct option is A $5(\sqrt{3}-1)$

$u_x=100\times cos{60}^{\circ }=50$
$u_y=100\times sin{60}^{\circ }=50\sqrt{3}$
$v_y=u_y+a_{y}t=50\sqrt{3}-gt$ and $v_x=u_x=50$
When angle is ${45}^{\circ }$
$tan{45}^{\circ }=\frac{v_y}{v_x}\Rightarrow v_y=v_x\Rightarrow 50\sqrt{3}-gt=50$
$\Rightarrow 50(\sqrt{3}-1)=gt\Rightarrow t=5(\sqrt{3}-1)s$