Motion in a Plane

A boy playing on the roof of a 10 m high building throws a ball with a speed of 10 m/s at an angle of $$30^\circ$$ with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground? $$\left[ {g = 10m/{s^2},\sin 30^\circ = {1 \over 2},\cos 30^\circ = {{\sqrt 3 } \over 2}} \right]$$

A projectile can have the same range 'R' for two angles of projection. If T₁ and T₂ be the time of flights in the two cases, then the product of the two time of flights is directly proportional to

A ball is thrown from a point with a speed ν₀ at an angle of projection θ. From the same point and at the same instant person starts running with a constant speed $${{{v_0}} \over 2}$$ to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection θ?

A particle is moving with velocity $$\overrightarrow v = k\left( {y\widehat i + x\widehat j} \right)$$, where K is a constant. The general equation for its path 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 :

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

A projectile is given an initial velocity of $$\left( {\widehat i + 2\widehat j} \right)$$ m/s, where $${\widehat i}$$ is along the ground and $${\widehat j}$$ is along the vertical. If g = 10 m/s², the equation of its trajectory is:

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 -

A shell is fired from a fixed artillery gun with an initial speed u such that it hits the target on the ground at a distance R from it. If t₁ and t₂ are the values of the time taken by it to hit the target in two possible ways, the product t₁t₂ is -

The trajectory of a projectile near the surface of the earth is given as y = 2x – 9x² . If it were launched at an angle $$\theta$$₀ with speed v₀ then (g = 10 ms^–2) :

Two particles are projected from the same point with the same speed u such that they have the same range R, but different maximum heights, h₁ and h₂. Which of the following is correct ?

The trajectory of a projectile in a vertical plane is y = $$\alpha$$x $$-$$ $$\beta$$x², where $$\alpha$$ and $$\beta$$ are constants and x & y are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection $$\theta$$ and the maximum height attained H are respectively given by :

A bomb is dropped by fighter plane flying horizontally. To an observer sitting in the plane, the trajectory of the bomb is a :

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 helicopter is flying horizontally with a speed 'v' at an altitude 'h' has to drop a food packet for a man on the ground. What is the distance of helicopter from the man when the food packet is dropped?

The ranges and heights for two projectiles projected with the same initial velocity at angles 42$$^\circ$$ and 48$$^\circ$$ with the horizontal are R₁, R₂ and H₁, H₂ respectively. Choose the correct option :

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 projectile is launched at an angle '$$\alpha$$' with the horizontal with a velocity 20 ms^$$-$$1. After 10 s, its inclination with horizontal is '$$\beta$$'. The value of tan$$\beta$$ will be : (g = 10 ms^$$-$$2).

Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Two identical balls A and B thrown with same velocity 'u' at two different angles with horizontal attained the same range R. IF A and B reached the maximum height h₁ and h₂ respectively, then $$R = 4\sqrt {{h_1}{h_2}}$$ Reason R : Product of said heights. $${h_1}{h_2} = \left( {{{{u^2}{{\sin }^2}\theta } \over {2g}}} \right)\,.\,\left( {{{{u^2}{{\cos }^2}\theta } \over {2g}}} \right)$$ Choose the correct answer :