Question

At what point on a projectile's trajectory, its speed is minimum? If a stone is thrown with a speed $y$, at an angle with horizontal, find the velocity of the stone when its line of motion makes an angle with horizontal, component of velocity will be?

Answer 1

In order to solve this question, we are going to first analyze the trajectory of a projectile and discuss the variation of speed and the point with the maximum speed is found. After that the component of the horizontal velocity is found at the point when its line of motion makes an angle with horizontal.

Complete step-by-step solution:
A projectile is projected making an angle with the horizontal, its horizontal speed remains the same while the vertical speed changes. The speed is maximum at the highest point. At the highest point of the trajectory of a projectile, its speed is minimum. This is because the horizontal speed of the projectile remains constant. Now, the only force acting on the projectile is gravity, and the acceleration is that due to gravity, so its speed is minimum at the highest point of the trajectory.
It is given that a stone is thrown with a velocity$y$, at an angle with horizontal
Then, the horizontal component of the velocity will be
${v_x} = y\cos \theta$
Where, $\theta$is the angle of the stone with the horizontal.
If after some time the velocity of the stone makes an angle, $\phi$with the horizontal, then the velocity will be
${v_x}' = y'\cos \phi$

Note: It is important to note that the Projectile motion is a form of motion experienced by a launched object. The horizontal component of the velocity remains constant throughout the trajectory while the vertical part keeps on changing.
At the highest point, $\phi = 0$
${y_x} = y\sin \phi = 0 = \min$