Question #0c86a

Question

Question #0c86a

1 Answer

Impact of this question

1752 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-21T09:02:48

Discussed below

Explanation:

Given

$h \to The height from which the canon ball was fired = 60 m$ $h->"The height from which the canon ball was fired"=60m$

$u \to The vlocity of projection = 25 \frac{m}{s}$ $u->"The vlocity of projection"=25m/s$

$α \to Angle of projection = 53^{\circ}$ $alpha -> "Angle of projection"=53^@$

$And {sin 53}^{\circ} = 0.8 and {cos 53}^{\circ} = 0.6$ $"And "sin53^@=0.8 and cos53^@=0.6$

Now

$Horizontal component of vel.of projection u {cos 53}^{\circ} = 25 \cdot 0.6 = 15 \frac{m}{s}$ $"Horizontal component of vel.of projection " ucos53^@=25*0.6=15m/s$

$Vertical component of vel.of projection u {sin 53}^{\circ} = 25 \cdot 0.8 = 20 \frac{m}{s}$ $"Vertical component of vel.of projection " usin53^@=25*0.8=20m/s$

At the time when it reaches at maximum height H from its point of projection its vertical component becomes zero, So considering $g = m s^{- 2}$ $g=ms^-2$ we get

$0^{2} = 20^{2} - 2 \cdot 10 \cdot H \Rightarrow H = 20 m$ $0^2=20^2-2*10*H=>H=20m$

Hence at its maximum height the total height fron ground
$H_{total} = 60 + 20 = 80 m$ $H_"total"=60+20=80m$

Now if its time of flight be t s
then it will undergo a vertical displacement downward during this time. So
-
$- 60 = 20 \cdot t - \frac{1}{2} \cdot 20 \cdot t^{2}$ $-60=20*t-1/2*20*t^2$

$\Rightarrow t^{2} - 2 t - 6 = 0$ $=>t^2-2t-6=0$

$\Rightarrow t = \frac{2 + \sqrt{4 - 4 \cdot (- 6) \cdot 1}}{2}$ $=>t=(2+sqrt(4-4*(-6)*1))/2$

$\Rightarrow t = (\sqrt{7} + 1) s$ $=>t=(sqrt7+1)s$

Net horizontal displacement during this time of flight

$= u cos α \cdot t = 15 (\sqrt{7} + 1) m$ $=ucosalpha*t=15(sqrt7+1)m$

Science

Math

Humanities

... and beyond

Question #0c86a

1 Answer

Explanation:

Related questions

Impact of this question