# Using Equations for Constant Acceleration- Vertical Motion of a Bullet

• L_0611
In summary, a bullet is shot vertically into the air at a speed of 512 m/s and reaches a maximum height of 26.1 m. It takes 52.2 seconds for the bullet to stop rising. After 60 seconds, the bullet's velocity is 76 m/s. However, it is important to note that this is only the speed and not the velocity, which includes both magnitude and direction.
L_0611

## Homework Statement

A bullet is shot vertically into the air at a speed of 512 m/s.

a) to what maximum height does the bullet go?

b) how much time passes before the bullet stops rising?

c) what is the velocity of the bullet after 60s?

b) a= (V2-V1)/Δt

c) V2=V1+aΔt

## The Attempt at a Solution

a) Given: V2: 0 m/s; V1: 512 m/s; a: 9.8 m/s²
Unknown: distance
Equation: d=(V2²+V1²)/2a
Substitute: d=(0 m/s + 512 m/s)/2(9.8 m/s²)
Solve 26.1 m
∴ the bullet will reach a maximum height of 26.1 m.

b) Given: V2: 0 m/s; V1: 512 m/s; a: 9.8 m/s²
Unknown: time
Equation: Δt=(V2-V1)/a
Substitute: Δt=(0 m/s-512 m/s)/9.8 m/s²
Solve: 52.2s
∴ the bullet will stop rising after 52.2s

c) Given: t: 60s; a: -9.8m/s²; V1: 512 m/s
Unknown: V2
Equation: V2=V1+aΔt
Substitute: V2=512 m/s + (-9.8 m/s²)(60s)
Solve: 76m/s
∴ the bullet's velocity after 60s is 76m/s

I would like for someone to please look over my work and tell me if it's correct or not. Thank you.

L_0611 said:

## Homework Statement

A bullet is shot vertically into the air at a speed of 512 m/s.

a) to what maximum height does the bullet go?

b) how much time passes before the bullet stops rising?

c) what is the velocity of the bullet after 60s?

b) a= (V2-V1)/Δt

c) V2=V1+aΔt

## The Attempt at a Solution

a) Given: V2: 0 m/s; V1: 512 m/s; a: 9.8 m/s²
Unknown: distance
Equation: d=(V2²+V1²)/2a
Substitute: d=(0 m/s + 512 m/s)/2(9.8 m/s²)
Solve 26.1 m
∴ the bullet will reach a maximum height of 26.1 m.
heck, i could throw a baseball that high at a mere fraction of that initial speed...you forgot to square it..
b) Given: V2: 0 m/s; V1: 512 m/s; a: 9.8 m/s²
Unknown: time
Equation: Δt=(V2-V1)/a
Substitute: Δt=(0 m/s-512 m/s)/9.8 m/s²
Solve: 52.2s
∴ the bullet will stop rising after 52.2s
looks good!
c) Given: t: 60s; a: -9.8m/s²; V1: 512 m/s
Unknown: V2
Equation: V2=V1+aΔt
Substitute: V2=512 m/s + (-9.8 m/s²)(60s)
Solve: 76m/s
∴ the bullet's velocity after 60s is 76m/s
that is its speed...what is its velocity?

## 1. What is the equation for calculating the vertical motion of a bullet with constant acceleration?

The equation for calculating the vertical motion of a bullet with constant acceleration is y = y0 + v0t + 1/2at2, where y is the final height, y0 is the initial height, v0 is the initial velocity, a is the acceleration, and t is the time.

## 2. How is the initial velocity of a bullet determined in the equation?

The initial velocity of a bullet is determined by measuring its speed at the moment it is fired from the gun or by using a chronograph to measure its speed as it leaves the barrel.

## 3. Can the equation for constant acceleration be used for a bullet that is fired at an angle?

Yes, the equation for constant acceleration can be used for a bullet that is fired at an angle, as long as the acceleration is constant and the motion is in the vertical direction.

## 4. What is the significance of the constant acceleration in the equation for vertical motion of a bullet?

The constant acceleration in the equation represents the constant force of gravity acting on the bullet, which causes it to accelerate towards the ground at a rate of 9.8 m/s2 (on Earth).

## 5. How can the equation for constant acceleration be used to predict the height reached by a bullet?

By using the equation y = y0 + v0t + 1/2at2 and plugging in the values for the initial height, initial velocity, and acceleration, the equation can be solved for the final height (or maximum height reached by the bullet). This can help in predicting the trajectory of the bullet and determining its range.

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