Explain how $u\cos \theta$ is the horizontal component and $u\sin \theta$ the vertical component when a projectile is projected at an angle $\theta$ from the horizontal.

Q. A body is projected with a velocity $u$ making an angle $\theta$ with the horizontal. The time after which its vertical component of velocity is equal to horizontal component is

Q. A projectile is thrown with a speed $u$ at an angle $\theta$ with the vertical. Its average velocity between the instants, it crosses half the maximum height is :

Q. Find the angle of projection of a projectile motion having same values of horizontal component and vertical component of velocity of projection.

Q. The vertical component of velocity of a projectile at its maximum height (u = velocity of projection, $\theta =$ angle of projection) is :

Q. Assertion :If a body is projected obliquely at angle $\theta$ above horizontal with initial speed $v$ then its speed at the instant when its velocity makes an angle $\alpha$ above the horizontal is $\frac{vcos\theta }{cos\alpha }$ Reason: Horizontal component of velocity of projectile remains constant.