Motion in a Plane

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation- y=O, u_y= V_ocos $\theta$

a_y=-gcos $\theta$ , t =T

applying equation of kinematics

y=u_yt+ $\frac{1}{2}{a}_y$ t²

0 = V_ocos $\theta T$ +T^{2 $\frac{1}{2}(-gcos\theta )$}

T= $\frac{2V_{o}cos\theta }{gcos\theta }$

T= 2V₀/g

X= L, u_x=V_osin $\theta$  , a_x= gsin $\theta$  , t=T= $\frac{2V_o}{g}$

X=u_xt+ $\frac{1}{2}{a}_x{t}^2$

L= V_osin $\theta T+\frac{1}{2}gsin\theta T^2$

L= $\frac{4v_o^2}{g}$ sin $\theta$

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation – particle is projected from the point O.

Let time taken in reaching from point O to point P is T.

for journey O to P

y=0,u_y= V_osin $\beta$ ,a_y= -gcos $\alpha ,t=T$

y=u_yt + $\frac{1}{2}{a}_y{t}^2$

0= V_osin $\beta T+\frac{1}{2}\left(-gcos\alpha \right)T^2$

T[V_osin $\beta -\frac{gcos\alpha }{2}$ T]=0

T = time of flight = $\frac{2V_{o}sin\beta }{gcos\alpha }$

 

Motion along OX

x= L

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation – target T is at horizontal distance x= R+ $?x$ and between point of projection y= -h

Maximum horizontal range R= $\frac{v_o^2}{g}\theta =45$ …………1

Horizontal component of initial velocity = V_ocos $\theta$

Vertical component of initial velocity = -V_osin $\theta$

So h

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation- speed of jackets = 125m/s

Height of hill = 500m

To cross the hill vertical component of velocity should be grater than this value u_y= $\sqrt[]{2gh}$

$=\sqrt[]{2*10*500}=100m/s$

So u²= u_x²+u_y²

Horizontal component of initial velocity u_x = $\sqrt[]{u^2-u_y^2}=\sqrt[]{{125}^2-{100}^2}=75m/s$

Time taken to reach the

4.32

(a) Let $\mathit{\theta }\left(\mathit{t}\right)=\mathit{}{\mathbf{tan}}^{-1}?\left(\frac{{\mathit{v}}_{\mathit{o}\mathit{y}}-\mathit{}\mathit{g}\mathit{t}}{{\mathit{v}}_{\mathit{o}\mathit{x}}}\right)$ be the angle at which the projectile is fired w.r.t. the x-axis, since ${\mathit{\theta }}_0=\mathit{}{\mathbf{tan}}^{-1}?\left(\frac{4{\mathit{h}}_{\mathit{m}}}{\mathit{R}}\right)$ depends on t

Therefore tan $\theta$ since (Vy=Voy -gt and Vx = Vox)

(b) Since $\theta$ = $\theta \left(t\right)=\frac{Vx}{Vy}=\frac{Voy-gt}{Vox}$ sin2 $\theta \left(t\right)={\tan }^{-1}?\left(\frac{Voy-gt}{Vox}\right)$ /2g…….(1)

R = $h_{max}$ sin2 $u^2$ /g …….(2)

Dividing (1) by (2)

( $\theta$ /R) =

4.31

Speed of the cyclist = 27 km/h = 7.5 m/s

Radius of the road = 80m

The net acceleration is due to braking and the centripetal acceleration

Acceleration due to braking = 0.5 m/s²

Centripetal acceleration a = v²r = (7.5)2/80= 0.703 m/s²

The resultant acceleration is given by a = sqrt ( $a_t^2$ + 

4.30

Speed of the fighter plane = 720 km/h = 200 m/s

The altitude of the plane = 1.5 km =1500 m

Velocity of the shell = 600 m/s

$\sin ?\theta$ = 200/600

$\therefore \theta =19.47°$

Let H be the minimum altitude for the plane to fly, without being hit

Using equation H = $u^2$ sin2 (90- $\theta )/2g$

= (6002cos2 $\theta$ )/2g

= 360000 { ( 1 + co

4.29 The angle of projectile $\theta$ = 30 $°$

The bullet hits the ground at a distance of 3 km, Range R = 3 km

We know horizontal range for a projectile motion, R = $u^2$ sin2 $\theta$ / g

3 = $u^2$ sin60 $°$ / g

$\frac{u^2}{g}$ = 3/ sin60 $°$ = 3.464 …………… (1)

To hit a target at 5 k