Time of Flight, Maximum Height & Range

If the initial velocity of projection be doubled, keeping the angle of projection same, the maximum height reached by it will

For a projectile motion, the range R is 'n' times the maximum height H. So the angle of projection is

A gun mounted on the ground fires bullets in all directions with same speed. The farthest distance the bullets could reach is 6.4 m. The speed of the bullets from the gun is $\underline{\qquad}\text{ m/s}$. (take $g = 10\text{ m/s}^{2}$)

A stone is thrown at an angle $\theta$ with the horizontal reaches a maximum height $H$, then the time of flight of stone will be ($g$ = acceleration due to gravity)

A boy throws a ball into air at $45^\circ$ from the horizontal to land it on a roof of a building of height $H$. If the ball attains maximum height in 2 s and lands on the building in 3 s after launch, then value of $H$ is ______ m.

$\left(g = 10\text{ m/s}^{2}\right)$

A body when projected at an angle ' $\theta$ ' with the horizontal reaches a maximum height ' $H$ '. The time of flight of the body will be ( $g =$ acceleration due to gravity)

The angle of projection of a particle is measured from the vertical axis as $\phi$ and the maximum height reached by the particle is $h_m$. Here $h_m$ as function of $\phi$ can be presented as

A particle is projected with velocity $u$ so that its horizontal range is three times the maximum height attained by it. The horizontal range of the projectile is given as $\frac{nu^2}{25g}$, where value of $n$ is: (Given, ' $g$ ' is the acceleration due to gravity.)

A particle is projected at an angle of $30^\circ$ from horizontal at a speed of $60\text{ m/s}$. The height traversed by the particle in the first second is $h_0$ and height traversed in the last second, before it reaches the maximum height, is $h_1$. The ratio $h_0 : h_1$ is __________.

[Take, $g = 10\text{ m/s}^{2}$ ]

The maximum height reached by a projectile is $64\text{ m}$. If the initial velocity is halved, the new maximum height of the projectile is ______ $\text{m}$.

For a projectile, the maximum height and horizontal range are same. The angle of projection ' $\theta$ ' of the projectile is

The angle of projection for a projectile to have same horizontal range and maximum height is:

A bullet is fired from a gun at the speed of $280\text{ ms}^{-1}$ in the direction $30^\circ$ above the horizontal. The maximum height attained by the bullet is

$\left(g = 9.8\text{ ms}^{-2},\ \sin 30^\circ = 0.5\right)$ :-

The maximum vertical height to which a man can throw a ball is 136 m. The maximum horizontal distance upto which he can throw the same ball is:

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R

Assertion A: When a body is projected at an angle $45^\circ$, it's range is maximum.

Reason R: For maximum range, the value of $\sin 2\theta$ should be equal to one.

In the light of the above statements, choose the correct answer from the options given below:

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:

A projectile fired at $30^\circ$ to the ground is observed to be at same height at time $3\text{ s}$ and $5\text{ s}$ after projection, during its flight. The speed of projection of the projectile is ___________ $\text{m s}^{-1}$. (Given $g = 10\text{ m s}^{-2}$)

A large number of bullets are fired in all directions with same speed '$U$'. The maximum area on the ground on which the bullets will spread is

A shell is fired at an angle of $30^\circ$ to the horizontal with velocity $196\text{ m/s}$. The time of flight is $\left[\sin 30^\circ = \frac{1}{2} = \cos 60^\circ\right]$

A cricket ball is thrown by a player at a speed of 20 m/s in a direction 30$^\circ$ above the horizontal. The maximum height attained by the ball during its motion is

(g = 10 m/s2)

A person can throw a ball upto a maximum range of 100 m. How high above the ground he can throw the same ball?

A ball is projected from the ground with a speed 15 ms${}^{-1}$ at an angle $\theta$ with horizontal so that its range and maximum height are equal,

then 'tan $\theta$' will be equal to:

A projectile is projected with velocity of 25 m/s at an angle $\theta$ with the horizontal. After t seconds its inclination with horizontal becomes zero. If R represents horizontal range of the projectile, the value of $\theta$ will be:

[use g = 10 m/s2]

A cricket player hit a ball like a projectile but the fielder caught the ball after 2 second. The maximum height reached by a ball is (g = 10 m/s2)

A player kicks a football with an initial speed of 25 ms${}^{-1}$ at an angle of 45$^\circ$ from the ground. What are the maximum height and the time taken by the football to reach at the highest point during motion? (Take g = 10 ms${}^{-2}$)

A particle moving in a circle of radius R with a uniform speed takes a 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:

Two guns A and B can fire bullets at speeds 1 km/s and 2 km/s respectively. From a point on a horizontal ground, they are fired in all possible directions. The ratio of maximum areas covered by the bullets fired by the two guns, on the ground is -

The initial speed of a bullet fired from a rifle is 630 m/s. The rifle is fired at the center of a target 700 m away at the same level as the target. How far above the center of the target the rifle must be aimed in order to hit the target?

A projectile is fired from the surface of the earth with a velocity of 5 m s−1 and angle $\theta$ with the horizontal. Another projectile fired from another planet with a velocity of 3 m s−1 at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet is (in m s${}^{-2}$) is

(Given g = 9.8 m s${}^{-2}$)

A boy can throw a stone up to a maximum height of 10 m. The maximum horizontal distance that the boy can throw the same stone up to will be

The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is

A missile is fired for maximum range with an initial velocity of 20 m/s. If g = 10 m/s2, the range of the missile is

A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is v, the total area around the fountain that gets wet is:

The maximum range of a gun of horizontal terrain is 16 km. If $g = 10\text{ m s}^{-2}$, then muzzle velocity of a shell must be