Using Equations for Constant Acceleration- Vertical Motion of a Bullet

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SUMMARY

The discussion focuses on the vertical motion of a bullet shot at an initial speed of 512 m/s. The maximum height reached by the bullet is calculated to be 26.1 meters using the equation V2²=V1²+2ad. The time taken for the bullet to stop rising is determined to be 52.2 seconds using the formula Δt=(V2-V1)/a. After 60 seconds, the bullet's velocity is found to be 76 m/s, calculated with V2=V1+aΔt, where acceleration due to gravity is considered as -9.8 m/s².

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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.
 
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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?

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.
heck, i could throw a baseball that high at a mere fraction of that initial speed...you forgot to square it..:wink:
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?
 

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