A projectile is fired with velocity v at angle thita. Find the time when velocity of projectile will make angle 90' with initial direction of velocity?

To find the time when the velocity of the projectile makes an angle of 90 degrees with the initial direction of velocity, we need to consider the motion of the projectile in both the horizontal and vertical directions separately.

Let's break down the initial velocity into its horizontal and vertical components. The horizontal component (Vx) of the initial velocity is given by Vx = v * cos(theta), and the vertical component (Vy) is given by Vy = v * sin(theta).

At the instant when the velocity makes an angle of 90 degrees with the initial direction, the vertical component of the velocity becomes zero. This happens at the highest point of the projectile's trajectory.

Now, let's analyze the vertical motion of the projectile. We can use the kinematic equation for vertical motion:

Vy = Vy0 + a * t

where Vy is the vertical velocity at time t, Vy0 is the initial vertical velocity (which is Vy), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time.

Since the velocity becomes zero at the highest point, we can substitute Vy = 0 into the equation:

0 = Vy0 - 9.8 * t

Solving for t, we have:

t = Vy0 / 9.8

Now, substituting the value of Vy0, we get:

t = (v * sin(theta)) / 9.8

Therefore, the time when the velocity of the projectile makes an angle of 90 degrees with the initial direction is given by t = (v * sin(theta)) / 9.8.