# Projectile Motion Problem: Finding Velocity Direction at Impact

• EHogeberg
In summary, a projectile is fired with an initial speed of 16.0 m/s at an angle of 45.0^\circ above the horizontal and hits the ground 9.00 seconds later. The magnitude of its velocity an instant before it hits the ground is 77.7 m/s and its maximum height is 301.53 m. Using the given equations, it can be determined that the angle may not have been exactly 45 degrees, as it would have hit the ground in just over 2 seconds. The vertical speed at the point of impact would have made the angle steeper.
EHogeberg

## Homework Statement

A projectile is fired with an initial speed of 16.0 m/s at an angle of 45.0^\circ above the horizontal. The object hits the ground 9.00 s later.

What is the direction (below +x) of the projectile's velocity at the instant it hits the ground?

Vo=16m/s
a=-g
θ=45
Tf=9 seconds
magnitude of velocity an instant before it hits the ground: 77.7 m/s
max height: 301.53 m/s

## Homework Equations

-gsin(θ)
Vxf=Vxi+a(tf-ti)
Xf=Xi+Vxi(tf-ti) + 1/2 A(tf-ti)^2
Vxf^2=Vxi^2+2a(Xf-Xi)

Vyf=Vyi+a(tf-ti)
Yf=yi+Vyi(tf-ti) + 1/2 A(tf-ti)^2
Vyf^2=Vyi^2+2a(Yf-Yi)

## The Attempt at a Solution

I got 45 degrees and I got that from finding the xcomponents and ycomponents

EHogeberg said:

## Homework Statement

A projectile is fired with an initial speed of 16.0 m/s at an angle of 45.0^\circ above the horizontal. The object hits the ground 9.00 s later.

What is the direction (below +x) of the projectile's velocity at the instant it hits the ground?

Vo=16m/s
a=-g
θ=45
Tf=9 seconds
magnitude of velocity an instant before it hits the ground: 77.7 m/s
max height: 301.53 m/s

## Homework Equations

-gsin(θ)
Vxf=Vxi+a(tf-ti)
Xf=Xi+Vxi(tf-ti) + 1/2 A(tf-ti)^2
Vxf^2=Vxi^2+2a(Xf-Xi)

Vyf=Vyi+a(tf-ti)
Yf=yi+Vyi(tf-ti) + 1/2 A(tf-ti)^2
Vyf^2=Vyi^2+2a(Yf-Yi)

## The Attempt at a Solution

I got 45 degrees and I got that from finding the xcomponents and ycomponents

If it took 9 seconds to hit the ground, it must have been fired from the top of a rather tall building/high hill.

At that speed, at 45 degrees it would hit the ground [flat ground scenario] in a little over 2 seconds.

In light of what PeterO has written, I'd check that 45 degrees angle. It would have been 45 degrees after 2 secs of flight, so if it took another 7 secs, then its vertical speed would have been considerably more, making the angle steeper.

## 1. What is projectile motion?

Projectile motion is the motion of an object that is launched into the air and moves along a curved path under the influence of gravity. This type of motion is a combination of horizontal and vertical motion.

## 2. What are the key components of projectile motion?

The key components of projectile motion are the initial velocity, launch angle, and acceleration due to gravity. The initial velocity is the speed at which the object is launched, the launch angle is the angle at which the object is launched, and the acceleration due to gravity is the force that pulls the object towards the ground.

## 3. How do you calculate the range of a projectile?

The range of a projectile can be calculated using the equation R = (v^2 * sin2θ) / g, where R is the range, v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity. This equation assumes that there is no air resistance.

## 4. What is the maximum height of a projectile?

The maximum height of a projectile is achieved at the highest point of its trajectory, where the vertical velocity is equal to 0. This can be calculated using the equation h = (v^2 * sin^2θ) / 2g, where h is the maximum height, v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

## 5. How does air resistance affect projectile motion?

Air resistance can affect projectile motion by slowing down the object's horizontal velocity and changing its trajectory. This means that the range and maximum height of the projectile will be reduced. Air resistance can also cause the object to land at a different angle than it was launched at.

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