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3-2.Motion in Plane
medium
The velocity of projection of a body is increased by $2 \% .$ Other factors remaining unchanged, what will be the percentage change in the maximum height attained ? (in $\%$)
A
$1$
B
$2$
C
$4$
D
$8$
(AIIMS-2019)
Solution
The formula for the maximum height during projectile motion is,
$h=\frac{(u \sin \theta)^{2}}{2 g}$
$h=\frac{u^{2} \sin ^{2} \theta}{2 g}$
Now,
$\frac{\Delta h}{h}=\frac{2 \Delta u}{u}$
$\frac{\Delta h}{h}=2 \times 2=4 \%$
Standard 11
Physics
Similar Questions
Column $-I$ Angle of projection |
Column $-II$ |
$A.$ $\theta \, = \,{45^o}$ | $1.$ $\frac{{{K_h}}}{{{K_i}}} = \frac{1}{4}$ |
$B.$ $\theta \, = \,{60^o}$ | $2.$ $\frac{{g{T^2}}}{R} = 8$ |
$C.$ $\theta \, = \,{30^o}$ | $3.$ $\frac{R}{H} = 4\sqrt 3 $ |
$D.$ $\theta \, = \,{\tan ^{ – 1}}\,4$ | $4.$ $\frac{R}{H} = 4$ |
$K_h :$ kinetic energy at the highest point
medium
hard