- #1

L_0611

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## 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?

## Homework Equations

a) V2²=V1²+2ad

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.