Tag Archives: Projectile Motion

IIT JEE

A projectile is thrown from a point O ….

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A projectile is thrown from a point O on the ground at an angle 45° from the vertical and with a speed 5√2 m/s. The projectile at the highest point of its trajectory splits into two equal parts. One part falls vertically down to the ground, 0.5 s after the splitting. The other part, t seconds after the splitting, falls to the ground at a distance x meters from the point O. The acceleration due to gravity $g = 10 m/s^2$.

Q.1 The value of t is ___ .
Q.2 The value of x is ___ .

Solution

The part that falls vertically down to the ground has initial speed = 0. So, it falls freely taking 0.5 s to reach the ground. Continue reading A projectile is thrown from a point O …. →

Physics

A particle moving in a circle of radius R ….

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A particle moving in a circle of radius R with a uniform speed takes time T to complete one revolution. If this particle were projected with the same speed at an angle ‘$\theta $’ to the horizontal, the maximum height attained by it equals 4R. The angle of projection ‘$\theta $’ is then given by:

(1) $\theta  = {\cos ^{ – 1}}\sqrt {\frac{{g{T^2}}}{{{\pi ^2}R}}} $
(2) $\theta  = {\cos ^{ – 1}}\sqrt {\frac{{{\pi ^2}R}}{{g{T^2}}}} $
(3) $\theta  = {\sin ^{ – 1}}\sqrt {\frac{{{\pi ^2}R}}{{g{T^2}}}} $
(4) $\theta  = {\sin ^{ – 1}}\sqrt {\frac{{2g{T^2}}}{{{\pi ^2}R}}} $ Continue reading A particle moving in a circle of radius R …. →

JEE Main

The ranges and heights for two projectiles projected ….

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The ranges and heights for two projectiles projected with the same initial velocity at angles $42^\circ $ and $48^\circ $ with the horizontal are $R_1\,,R_2$ and $H_1\,,H_2$ respectively. Choose the correct option:

(A) $R_1 < R_2 $ and $H_1 < H_2 $
(B) $R_1 > R_2 $ and $H_1 = H_2 $
(C) $R_1 = R_2 $ and $H_1 = H_2 $
(D) $R_1 = R_2 $ and $H_1 < H_2 $

Solution

$R=\frac {u^2 sin 2\theta }{g} $, $H=\frac {u^2 sin^2 \theta }{2g}$

$\frac {R_1}{R_2} = \frac {sin 2 \times 42^\circ}{sin 2\times 48^\circ} = \frac {cos 6^\circ}{cos 6^\circ} = 1 $

$\frac {H_1}{H_2}= \frac {sin^2 42^\circ }{sin^2 48^\circ } = tan^2 42^\circ < 1 $

Answer: (D)