The range of a particle when launched at an angle of ${15^o}$ with the horizontal is $1.5 \,km$. What is the range of the projectile when launched at an angle of ${45^o}$ to the horizontal ........ $km$
$1.5 $
$3.0$
$6.0$
$0.75$
The initial speed of a projectile fired from ground is $u$. At the highest point during its motion, the speed of projectile is $\frac{\sqrt{3}}{2} u$. The time of flight of the projectile is:
Given that $u_x=$ horizontal component of initial velocity of a projectile, $u_y=$ vertical component of initial velocity, $R=$ horizontal range, $T=$ time of flight and $H=$ maximum height of projectile. Now match the following two columns.
Column $I$ | Column $II$ |
$(A)$ $u_x$ is doubled, $u_y$ is halved | $(p)$ $H$ will remain unchanged |
$(B)$ $u_y$ is doubled $u_x$ is halved | $(q)$ $R$ will remain unchanged |
$(C)$ $u_x$ and $u_y$ both are doubled | $(r)$ $R$ will become four times |
$(D)$ Only $u_y$ is doubled | $(s)$ $H$ will become four times |
The horizontal range and the maximum height of a projectile are equal . The angle of projection of the projectile is
The equation of motion of a projectile is $y=12 x-\frac{3}{4} x^2$ $..........\,m$ is the range of the projectile.
A particle is thrown with a velocity of $u \,m / s$. It passes $A$ and $B$ as shown in figure at time $t_1=1 \,s$ and $t_2=3 \,s$. The value of $u$ is ....... $m / s$ $\left(g=10 \,m / s ^2\right)$