Question

at what angel should a body be projected with a velocity 20/sec just to pass over the obstalace 12m high at a horizontal distance of 24m ?take g= 10m/sec​

Answer 1

Answer:

$\boxed{\mathfrak{Angle \ of \ projection \ (\theta) = {tan}^{ - 1} (2) \: or \: = 53 \degree}}$

Given:

Initial velocity of projectile (u) = 20 m/s

Horizontal distance (x) = 24 m

Vertical distance (y) = 12 m

Accelration due to gravity (g) = 10 m/s

Explanation:

Equation of trajectory:

$\boxed{ \bold{y = xtan \theta - \frac{g {x}^{2} }{2 {u}^{2} {cos}^{2} \theta } }}$

θ → Angle of projection

By substituting values in the equation we get:

$\rm \implies 12 = 24 \: tan \theta - \dfrac{10 \times {(24)}^{2} }{2 \times {(20)}^{2} \times {cos}^{2} \theta} \\ \\ \rm \implies 12 = 24 \: tan \theta - \dfrac{5760}{800 \: {cos}^{2} \theta} \\ \\ \rm \implies 12 = 24 \:tan \theta - \dfrac{36 \: {sec}^{2} \theta }{5} \\ \\ \rm \implies (12 = 24 \:tan \theta - \dfrac{36 \: {sec}^{2} \theta }{5} ) \times 5 \\ \\ \rm \implies 60= 120 \:tan \theta - 36 \: {sec}^{2} \theta \\ \\ \rm {sec}^{2} \theta = 1 + {tan}^{2} \theta : \\ \rm \implies 60 = 120 \:tan \theta - 36(1 + {tan}^{2} \theta) \\ \\ \rm \implies 60 = 120 \:tan \theta - 36 - 36 {tan}^{2} \theta) \\ \\ \rm \implies 36 \: {tan}^{2} \theta - 120 \: tan \theta + 96 = 0 \\ \\ \rm \implies 12(3 \: {tan}^{2} \theta - 10 \: tan \theta + 8) = 0 \\ \\ \rm \implies 3 \: {tan}^{2} \theta - 10 \: tan \theta + 8 = 0 \\ \\ \rm \implies 3 \: {tan}^{2} \theta - 6 \: tan \theta - 4\: tan \theta+ 8 = 0 \\ \\ \rm \implies 3 \: tan \theta(tan \theta - 2) - 4(tan \theta - 2) = 0 \\ \\ \rm \implies (tan \theta - 2)(3 \: tan \theta - 4) = 0 \\ \\ \rm \implies tan \theta = 2 \: \: \: \: or \: \: \: \: tan \theta = \frac{4}{3} \\ \\ \rm \implies \theta = {tan}^{ - 1} (2) \: \: \: \: or \: \: \: \: \theta = {tan}^{ - 1} ( \dfrac{4}{3} ) \\ \\ \rm \implies \theta = {tan}^{ - 1} (2) \: \: \: \: or \: \: \: \: \theta = 53 \degree$

Answer 2

Taking horizotal motion of body projected from O from O to Awehave,u_(x) =u cos theta =20 cos theta ms^-1),x_(0) =0, x=24 m, a_(x) =0, t=t (say)Asx=x_(0) +u_(x) t + 1/2 a_(x) t_(2)∴24 =0 + ( u cos theta) xx t =(20 cos thera t)ort=(24)/( 20 cos theta) =6/(5 cos theta)Tak∈gverticalupwardmotionofbodyform(O)→(A),

we have : yy=usinθ=20sinθ,y0=0,

y=12m,ay=−10ms−2,t=t

As, y=y0+uyt+12ayt2

:. 12=0+(20sinθ)t+12×(−10)t2

or 12=(20sinθ(20sinθ)×65cosθ5×(65cosθ)2

or 12=24tanθ=5×3625cos2θ

=24tanθ(3605sec2θ

or 12=24tanθ−365(1+tan2θ)or36 tan ^(2) -120 tan theta + 96 =0or3 tan^(2) theta -10 tan theta + 8 =0or (tan theta -2) (3 tan theta -4) =0ortanθ=2or4/3

or θ63∘26'or53∘4'.