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Horizontally Launched Projectile - Angle of Velocity Just Before Hitting Ground

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Given the initial height $h$, the initial distance the projectile is above the ground, and the horizontal distance $d$ between the projectile's initial and final positions (from launch position to when it hits the ground), how do we find the angle of the a horizontally launched projectile's velocity vector just before it hits the ground? Since we are not given the initial velocity that the projectile is launched at, we can see that since it was launched horizontally, there is no vertical component of the initial velocity. So, we can see that since velocity $v = \frac{\Delta x}{t}$ and $d=\Delta x$, we have that $v_0 = \frac{d}{t}$. This initial horizontal velocity is constant for the projectile since no horizontal forces are acting on the projectile (we are assuming that air resistance is negligible), so we have that $v_x = v_0 = \frac{d}{t}$. Now, we notice that we are asked to find the angle $\theta$ of the projectile's final velocity. Since we know that the final vel...
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Van Allen Radiation Belts

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Before we dive into the Van Allen Radiation Belt, let's start with solar wind. The Sun constantly ejects plasma, mostly consisting of protons and electrons, out into space known as solar wind . Some solar wind carrying charged particles reaches Earth. The intensity of the solar wind varies depending on the temperature, velocity, and density from different regions of the Sun it is ejected from. Although most solar wind that reaches Earth is slow speed wind, high speed solar wind and powerful bursts of plasma from the sun, known as coronal mass ejections (CMEs) can cause damage. The immense radiation in the charged particles of solar wind is dangerous to satellites and would be especially harmful if it reached Earth's surface. Thankfully, we have Earth's magnetic field and the Van Allen Radiation Belts that protect us from solar wind and even more powerful coronal mass ejections. The Van Allen Radiation Belt consists of energetic charged particles, mostly from solar wind, t...
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Coefficient of Friction on an Inclined Surface

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An interesting relationship between the coefficient of kinetic friction $\mu_k$ between a wood block and an inclined surface and the angle of incline when the block travels at a constant speed down the incline. The key requirement for this scenario is to have the block travel at a constant speed down the incline, which means that the acceleration is equal to zero . If the acceleration is zero, that means the net force  $F_{net} = ma$ is also zero. So, we can draw out the forces with a free-body diagram as in the diagram below: We have three forces acting on the block as it slides down the incline: $F_g$, the force of gravity which pulls the block toward the Earth (or ground); $F_N$, the normal force which acts on the block perpendicular to the incline; and $F_{fr}$, the force of kinetic friction which acts opposite the block's direction of motion. Since the net force on the block is zero, the force of gravity parallel to the incline must equal the force of friction and the f...
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