MCQ
A projectile has same range R when the maximum height attained by it is either $h _1$ or $h _2$, Then R, hi and $h _2$, will be related as
- A$R=\sqrt{h_1 h_2}$
- B$R=2 \sqrt{h_1 h_2}$
- C$R =3 \sqrt{h_1 h_2}$
- ✓$R=4 \sqrt{h_1 h_2}$
✓
Answer
Correct option: D.
$R=4 \sqrt{h_1 h_2}$
(D)
The horizontal range is same for two angles of projection $\theta^{\circ}$ and $(9 0 -\theta)^{\circ}$
For $\theta^{\circ}, h _1=\frac{ u ^2 \sin ^2 \theta}{2 g}$ and For $(90-\theta)^{\circ}, h _2=\frac{ u ^2 \sin ^2(90-\theta)}{2 g}=\frac{ u ^2 \cos ^2 \theta}{2 g}$
$h _1 h_2=\frac{ u ^4}{4 g^2} \sin ^2 \theta \cos ^2 \theta$
$=\frac{1}{16}\left[\frac{2 u ^2 \sin \theta \cos \theta}{g}\right]^2$
$=\frac{1}{16}\left[\frac{ u ^2 \sin 2 \theta}{g}\right]^2$
$h _1 h_2=\frac{ R ^2}{16}$
$\therefore \quad R=4 \sqrt{h_1 h_2}$
The horizontal range is same for two angles of projection $\theta^{\circ}$ and $(9 0 -\theta)^{\circ}$
For $\theta^{\circ}, h _1=\frac{ u ^2 \sin ^2 \theta}{2 g}$ and For $(90-\theta)^{\circ}, h _2=\frac{ u ^2 \sin ^2(90-\theta)}{2 g}=\frac{ u ^2 \cos ^2 \theta}{2 g}$
$h _1 h_2=\frac{ u ^4}{4 g^2} \sin ^2 \theta \cos ^2 \theta$
$=\frac{1}{16}\left[\frac{2 u ^2 \sin \theta \cos \theta}{g}\right]^2$
$=\frac{1}{16}\left[\frac{ u ^2 \sin 2 \theta}{g}\right]^2$
$h _1 h_2=\frac{ R ^2}{16}$
$\therefore \quad R=4 \sqrt{h_1 h_2}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Transmission lines start radiating
The kinetic energy of a revolving satellite (mass $m$ ) at a height equal to thrice the radius of the earth (R) is
→Two identical wires of substance ' $P$ ' and ' $Q$ ' are subjected to equal stretching force along the length. If the elongation of ' $Q$ ' is more than that of ' $P$ ', then
→The velocity-time graph of a body is shown in the following graph At point C
If gravitational force of Earth disappears, what will happen to the satellite revolving round the Earth?
A particle $P$ starts from the origin with velocity $\overrightarrow{ u }=(2 \hat{ i }-4 \hat{ j }) m / s$ with constant acceleration $(3 \hat{ i }+5 \hat{ j }) m / s ^2$. After travelling for 2 seconds, its distance from the origin is
→A cyclist comes to skidding spot in 10 m. If the opposing force on the cycle due to the road is 200 N. The work done by the road on the cycle is
The bob of a conical pendulum undergoes ___________
Human heart is pumping blood with constant velocity $v m s ^{-1}$ at the rate of $M kg s ^{-1}$.The force required for this is (in N)
→When the observer moves towards the stationary source with velocity, $V_1$, the apparent frequency of emitted note is $F_1$. When the observer moves away from the source with velocity $V_1$, the apparent frequency is $F_2$. If $V$ is the velocity of sound in air and $\frac{F_1}{F_2}=2$ then $\frac{V}{V_1}=$ ?
→