Tag Archives: Projectile Motion
A projectile is thrown from a point O ….
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 ….
A particle moving in a circle of radius R ….
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 ….
The ranges and heights for two projectiles projected ….
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)
A projectile is given an initial velocity of $$\hat i + 2\hat j$$
A projectile is given an initial velocity of 
, where 
is along the ground and 
is along the vertical. If g = 10 m/s2, the equation of its trajectory is:
(1) 
(2) 
(3) 
(4) 
[JEE Main 2013]
Continue reading A projectile is given an initial velocity of $$\hat i + 2\hat j$$