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Model Paper 4 — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsModel Paper 45 Marks
Question
A projectile is projected horizontally with a velocity $u$. Show that its trajectory is parabolic. And obtain expressions for:
i. Time of flight
ii. Horizontal range
iii. Velocity at any instant $t$.
✓

Answer

Projectile fired parallel to horizontal. As shown in figure, suppose a body is projected horizontally with velocity u from a point O at a certain height h above the ground level. The body is under the influence of two simultaneous independent motions:
i. Uniform horizontal velocity u.
ii. Vertically downward accelerated motion with constant acceleration g.
Under the combined effect of the above two motions, the body moves along the path OPA.
Image
Horizontal projection of a projectile.
Trajectory of the projectile. After the time t , suppose the body reaches the point $P ( x , y )$. The horizontal distance covered by the body in time $t$ is
$
x=ut . \therefore t=\frac{x}{u}
$
The vertical distance travelled by the body in time $t$ is given by
$
s=u t+\frac{1}{2} a t^2
$
or $y =0 \times t+\frac{1}{2} g t^2=\frac{1}{2} g t^2$ [For vertical motion, $u =0$ ]
or $y=\frac{1}{2} g\left(\frac{x}{u}\right)^2=\left(\frac{g}{2 u^2}\right) x^2\left[\because t=\frac{x}{u}\right]$
or $y = kx ^2$ [Here $k =\frac{g}{2 u^2}=$ a constant]
As y is a quadratic function of x , so the trajectory of the projectile is a parabola.
Time of flight. It is the total time for which the projectile remains in its flight (from O to A ). Let T be its time of flight.
For the vertical downward motion of the body, we use
$
s=u t+\frac{1}{2} a t^2 \text { or h }==0 \times T+\frac{1}{2} g T^2
$
or $T-\sqrt{\frac{2 h}{g}}$
Horizontal range. It is the horizontal distance covered by the projectile during its time of flight. It is equal to $O A=R$. Thus
$R =$ Horizontal velocity $\times$ time of flight $= u \times T$
or $R =u \sqrt{\frac{2 h}{g}}$
Velocity of the projectile at any instant. At the instant $t$ (when the body is at point P ), let the velocity of the projectile be v . The velocity v has two rectangular components:
Horizontal component of velocity, $v _{ x }= u$
Vertical component of velocity, $v _{ y }=0+ gt = gt$
$\therefore$ The resultant velocity at point P is
$
v=\sqrt{v_x^2+v_y^2}=\sqrt{u^2+g^2 t^2}
$
If the velocity v makes an angle p with the horizontal, then $\tan \beta=\frac{v_y}{v_x}=\frac{g t}{u}$
or $\beta=\tan ^{-1}\left(\frac{g t}{u}\right)$

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