Using Equations for Constant Acceleration- Vertical Motion of a Bullet

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The discussion focuses on the calculations related to a bullet shot vertically at 512 m/s. The maximum height reached by the bullet is calculated to be 26.1 m, with the bullet taking approximately 52.2 seconds to stop rising. After 60 seconds, the bullet's velocity is determined to be 76 m/s. There are some corrections noted, such as the need to square the initial velocity for height calculations and clarifications regarding the distinction between speed and velocity. Overall, the calculations are mostly correct but require careful attention to detail.
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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