Projectile Mathematics

Today I want to have some physics in the blog, something I've been neglecting for a long time. The question is about projectiles, walls, velocities, and directions.

A particle is projected 15m above the ground, and it just clears a wall 26.25m high and 30m away. What was the velocity and direction of the particle when it was projected?

This situation calls for some variables, given information, and physics formulas. I'm going to use the gravitational constant, g, as 10 m/s².

Since the particle is projected 15m above the ground and reaches a height of 26.25m, it only rises 11.25m. Let's call this number h.

When a particle is projected, its velocity, v, is split into two velocities - a horizontal velocity (v_x) and a vertical velocity (v_y). Since air friction is ignored, v_x remains constant and v_y changes according to gravity.

First, to find v_y, we can use the law of conservation of energy, which shows how kinetic energy (at the moment of projection) turns into potential energy (when it clears the wall). The formula is:
mv_y²/2 = mgh

We can make 15m our starting height, so h = 11.25. Also, we can cancel m on both sides. The only variable left is v_y:
v_y²/2 = 10 * 11.25
v_y²/2 = 112.5
v_y² = 225
v_y = 15

The positive number is taken because the particle is projected upwards.

Now we have the vertical velocity at the time of projection. When the particle clears the wall, its vertical velocity is zero. Let's see how much times it takes the particle to get to this stage by the formula v = v₀ + at:
0 = 15 - 10t
10t = 15
t = 1.5 seconds

It takes the particle 1.5 seconds from projection until it clears the wall. Now we can find the horizontal velociy of the particle, using the formula x = x₀ + vt:
30 = 0 + 1.5v_x
v_x = 20 m/s

Now we're left with the direction. When you break the velocity into its components, v_x and v_y, you see that the tangent of the angle of projection is v_y/v_x. To find that angle:
tan θ = 15/20
tan θ = 3/4
θ ≈ 36.87°

Enjoy,
Nadav

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