The shorter the hang time, the smaller the initial value of vy. The projectile fired at a 30 degree angle has the least vy and hence the shortest hang time.

- At what angle can a projectile achieve the shortest distance?
- What is the initial velocity of a projectile?
- At what point in its trajectory does a projectile reach its maximum speed?
- How many seconds does the projectile take to reach its maximum height?
- How far below an initial straight path will a projectile fall in one second?
- Can you calculate the time taken for a projectile motion?
- At what position is the speed of the projectile the least?
- What is the minimum speed of an oblique projectile?

Vertical motion totally determines the time for projectile motion. As a result, any projectile with an initial vertical velocity of 21.2 m/s and a landing height of **10.0 m** below its beginning altitude spends 3.79 s in the air. The horizontal velocity remains constant during this time.

The horizontal velocity gives us **the actual distance** traveled by the projectile. In **this case**, that's 63.6 m. So the total distance traveled by the bullet is 93.6 m.

The speed of the bullet is therefore equal to 93.6 m divided by 3.79 s, or 24.5 m/s.

As you can see, if you know **the initial velocity**, then you can find **the final velocity** after some time has passed. In this case, we knew it took 3.79 s to reach the ground so we could calculate the final velocity immediately after shooting the gun. If you don't know how long it takes the projectile to land, then you can use **this equation**:

Final velocity = Initial velocity + Distance traveled per second * Time elapsed

In this case, we know the distance traveled is 93.6 m so we can substitute that into the equation to get: Final velocity = 21.2 + (93.6 * 1.02) = 44.4 m/s

The projectile on a trajectory would have a minimum velocity at the trajectory point (at the maximum height). Maximum velocity is possible at the projectile's initial projection point (due to initial acceleration) and when it impacts the earth. After that, it will be moving under gravity only.

Thus, the projectile reaches its maximum speed over distance X from launch.

X equals the vertical distance traveled by the projectile plus its horizontal distance from the origin to the point of impact.

For example, if the projectile is launched from the top of **a 100-foot tower** and lands in a pond 40 feet from the base of the tower, then X = 40 + 100 = 140 feet.

Therefore, the projectile reaches maximum speed 140 feet from launch.

1 Response from an Expert At 18 seconds, the missile has reached its peak. It takes up to 2 more seconds for the missile to start **its descent** and hit the ground.

As a result, if a projectile falls from **a specific height** for one second, it will have gone just 9.8 m. More generally, if it falls from **a height h** in one second, then its final distance will be 9.8 m h.

This means that if you fire a bullet at a tree and count how long it takes to hit the ground, you can calculate how high the tree was by looking at how far the bullet went. This is called "trajectory analysis" and is used by scientists who want to know how much rain or snow will fall in **different parts** of the country, where clouds come from, and many other things about **weather systems**.

Here's another example: A meteorologist might want to know how high a cloud is so he/she can estimate how much moisture there is inside it. They do this by shooting bullets at the cloud and measuring how far they go - the higher the better. Then they use information on how long it takes bullets to fall to calculate their height from the surface of Earth.

Finally, here's something really interesting related to trajectory analysis: If you shoot two bullets at the same target but at different times, you can calculate the speed of each bullet by counting how long it takes them to reach the ground.

Calculate how long it takes the missile to reach **its maximum height**. Apply the formula (0-V)/-32.2 ft/s2 = T, where V is the starting vertical velocity calculated in step 2. It would take 0.998 seconds if your initial vertical velocity was 32.14 ft/s. The unit ft/s2 stands for feet per second squared. -32.2 is the coefficient for converting feet to meters. Multiply 0 by 0.998 to get the estimated time in seconds.

At the highest point of its trajectory, a bullet has **the slowest speed**. This is due to the projectile's constant horizontal speed. Gravity, which is 9.8 m/s2 acting downwards and has no influence on the horizontal velocity, is the sole force acting on the projectile. At this point, the kinetic energy of the projectile is maximum.

The higher the muzzle velocity, the farther the projectile will travel before it hits the ground.

Projectiles in space do not suffer from gravity and can reach high speeds after initial launch. However, they are also far away from the earth at the end of their journey, so its hard to achieve **great distances** with little effort.

On earth, firearms use gunpowder for propulsion. In space, there is no such thing as gunpowder, but there are other ways to generate thrust. One method used by astronauts is to fire small solid-fuel rockets called motor pods at their vehicle. The fuel in **these pods** burns for about three minutes giving them **enough energy** to reach orbit.

Another method involves using a large chemical propellant tank as the engine of a rocket. This method produces much more power than the other option, but cannot be fired once it is attached to the rocket body. It must instead be refilled while in flight with something called an unpressurized transfer tank system (or UTT).

An oblique projectile's lowest and maximum velocity are 10 m/s and 20 m/s, respectively. Therefore its average velocity is 15 m/s.