A projectile is fired from the origin (at y = 0 m) as shown in the diagram. The initial velocity components are and The projectile reaches maximum height at point P, then it falls and strikes the ground at point Q, which is 20 m below the launch point. Assuming air resistance is negligible, what is the horizontal component of the velocity ( V x) of the projectile at point P?

Question

A projectile is fired from the origin (at y = 0 m) as shown in the diagram. The initial velocity components are and The projectile reaches maximum height at point P, then it falls and strikes the ground at point Q, which is 20 m below the launch point. Assuming air resistance is negligible, what is the horizontal component of the velocity ( V x) of the projectile at point P?

Answer 1

something is missing here.

Answer 2

A projectile is launched at a certain angle. After 4s, it hits the top of a building 500 m away. The height of the building is 50 m (assume g=10 m/s^2). the projectile was launched at angle of?

Answer 3

To find the horizontal component of the velocity (Vx) at point P, we need to consider the motion of the projectile in the horizontal direction. Since there is no force acting horizontally (assuming air resistance is negligible), the horizontal velocity will remain constant throughout the projectile's motion.

We are given the initial velocity components of the projectile, given by V0x and V0y. To find Vx at point P, we need to find the time it takes for the projectile to reach point P.

We can use the fact that the initial vertical velocity component is zero at the highest point of the projectile's trajectory. We can use the kinematic equation:

Vf^2 = Vi^2 + 2as

where Vf is the final velocity (which is zero at point P), Vi is the initial velocity, a is the acceleration (due to gravity, which is -9.8 m/s^2), and s is the displacement. In this case, the displacement is the height of point P, which we'll call h.

0 = (V0y)^2 + 2(-9.8)h

Solving for h gives:

h = (V0y)^2 / (2 * 9.8)

Once we find the height of point P, we can use the fact that Viy = V0y at the highest point of the trajectory. The time it takes to reach point P can be found using the kinematic equation:

Vfy = Viy + a * t

where Vfy is the final vertical velocity (which is zero at point P), and we solve for t:

0 = V0y + (-9.8) * t

Solving for t gives:

t = V0y / 9.8

Now that we have the time it takes to reach point P, we can find the horizontal displacement using the formula:

s = V0x * t

Therefore, the horizontal component of the velocity (Vx) at point P is given by:

Vx = s / t = V0x * t / t = V0x

So, the horizontal component of the velocity at point P is equal to the initial horizontal velocity component (V0x).