A particle is projected horizontally with a speed of Um/s from a point 130m above a horizontal plane. The particle hits the plane at a point which is at a horizontal distance of 80m away from the starting point. Find initial speed?

To solve the problem of a particle projected horizontally from a height, we need to analyze the motion in two dimensions: horizontal and vertical. The key here is to understand that the horizontal and vertical motions are independent of each other, and we can use the equations of motion to find the initial speed.

Understanding the Vertical Motion

The particle is projected from a height of 130 meters. Since it is projected horizontally, its initial vertical velocity is 0 m/s. We can use the following kinematic equation to find the time it takes for the particle to hit the ground:

• Vertical Distance (h): 130 m
• Initial Vertical Velocity (u): 0 m/s
• Acceleration (a): 9.81 m/s² (acceleration due to gravity)

The equation we will use is:

h = ut + (1/2)at²

Substituting the known values:

130 = 0 * t + (1/2) * 9.81 * t²

This simplifies to:

130 = 4.905t²

Now, solving for t:

t² = 130 / 4.905

t² ≈ 26.5

t ≈ √26.5 ≈ 5.15 seconds

Analyzing the Horizontal Motion

Next, we consider the horizontal motion. The horizontal distance covered by the particle is 80 meters. The horizontal speed (U) remains constant since there is no horizontal acceleration. We can use the formula:

Horizontal Distance = Horizontal Speed × Time

Substituting the known values:

80 = U * 5.15

Now, solving for U:

U = 80 / 5.15

U ≈ 15.5 m/s

Final Result

Thus, the initial speed of the particle projected horizontally is approximately 15.5 m/s.

This problem illustrates the independence of horizontal and vertical motions in projectile motion, allowing us to solve for unknowns using basic kinematic equations.