Question

A projectile is thrown in the upward direction making an angle of 60∘ with the horizontal direction with a velocity of 150ms−1. Then the time after which its inclination with the horizontal is 45∘ is

Answer 1

Answer: 9.31 s

Given:

$\sf Angle\:of\:projection(\theta)=60^{\circ}$

$\sf Horizontal\:velocity(u)=150\ m/s$

To find:

$\sf Time(t)\:after\:which\:its\:inclination\:with\:horizontal\:is\:45^{\circ}.$

Explanation:

$\sf Let\:its\:final\:velocity=v\:at\:\phi=45^{\circ}.\:$

$\sf We\:know\:that,$

$\sf Horizontal\:component\:of\:velocity\:remains\:constant\:during\:the\:projectile.$

$\sf \therefore \boxed{\sf v\:cos\phi=u\:cos\theta}$

$\Rightarrow \sf v\:cos\:45^{\circ}=u\:cos\:60^{\circ}$

$\Rightarrow \sf v\times\dfrac{1}{\sqrt{2} } =u\times\dfrac{1}{2}$

$\Rightarrow \sf v\times\dfrac{1}{\sqrt{2} } =150\times\dfrac{1}{2}$

$\Rightarrow \boxed{\sf v = \dfrac{150}{\sqrt{2}}\:m/s}$

$\sf For\:vertical\:motion,\:we\:know\:that,$

$\boxed{\sf v\:sin\phi=u\:sin\theta-gt}$

$\Rightarrow \sf v\:sin\:45^{\circ}=u\:sin\:60^{\circ}-9.8t$

$\Rightarrow \sf v\times \dfrac{1}{\sqrt{2}} =v\times\dfrac{\sqrt{3} }{2} -9.8t$

$\Rightarrow \sf \dfrac{150}{\sqrt{2}} \times \dfrac{1}{\sqrt{2}} = {150}\times\dfrac{\sqrt{3} }{2} -9.8t$

$\Rightarrow \sf \dfrac{150}{{2}} = {150}\times\dfrac{\sqrt{3} }{2} -9.8t$

$\Rightarrow \sf 9.8t= \dfrac{150\sqrt{3} -150}{2}$

$\Rightarrow \sf 9.8t= \dfrac{150}{2}(\sqrt{3}-1)$

$\Rightarrow \sf 9.8t= 125(1.73-1)$

$\Rightarrow \sf 9.8t= 91.25$

$\Rightarrow \boxed{\sf t= 9.31\:s}$

Hence, the time after which its inclination with the horizontal is 45° is 9.31 s.

Answer 2

At the two points of the trajectory during projectile motion, the horizontal component of the velocity is same. Then

150×12=v×12–√150×12=v×12 or v=1502–√ms−1v=1502ms-1

Initially : uy=usin60∘=1503–√2ms−1uy=usin60∘=15032ms-1

Finally : vy=vsin45∘=1502–√×12–√=1502ms−1vy=vsin45∘=1502×12=1502ms-1

But vy=uy+aytvy=uy+ayt or 1502=1503–√2−10t1502=15032-10t

10t=1502(