projectile motion?

[Addallat]

I'm spacing on this question

If a man jumps 2 meters into the air, with what vertical velocity does he leave the ground?

A. 5.2 m/s
b. 6.3 m/s
c. 7.0 m/s
d. 8.1 m/s


y = velocity initial (t) + 1/2 (-g) t^2

initial velocity = 0?

sqrt (2y/g) ~~~> sqrt (2/5) = time

v = d/t

this just doesn't seem like the easiest way about going about this problem.... some insights please thank you

---

Members don't see this ad.

 

[krogers21]

I'm spacing on this question

If a man jumps 2 meters into the air, with what vertical velocity does he leave the ground?

A. 5.2 m/s
b. 6.3 m/s
c. 7.0 m/s
d. 8.1 m/s


y = velocity initial (t) + 1/2 (-g) t^2

initial velocity = 0?

sqrt (2y/g) ~~~> sqrt (2/5) = time

v = d/t

this just doesn't seem like the easiest way about going about this problem.... some insights please thank you

V^2=Vi^2 +2a(delta y)
At 2 meters, v = 0
Vi^2=2(g)(2)
Vi=sqrt( 9.8*4)=sqrt(40)

Sqrt (36)= 6
Sqrt(40)= a bit above 6 and a bit below 7.
Sqrt (49)=7

Answer is B.
Correct?

 

[Addallat]

ty the answer is B

 

[PlumbLine]

Regarding your question about whether it's the easiest way to go about solving, I usually approach these using conservation of energy principles. It often ends up being similar math, but can provide shortcuts such as for this problem.

For example, this is just a Kinetic -> potential energy problem. At max height, total energy= potential energy. P.E.= mgh, which was converted by his starting kinetic energy, 1/2 mv^2. Set those equal to each other and you can cancel out mass. So you have gh = 1/2 v^2. This avoids having to calculate Time as you did in your approach. Just always remember the relationships of kinetic and potential energies and you can often save time.

 

[Addallat]

Regarding your question about whether it's the easiest way to go about solving, I usually approach these using conservation of energy principles. It often ends up being similar math, but can provide shortcuts such as for this problem.

For example, this is just a Kinetic -> potential energy problem. At max height, total energy= potential energy. P.E.= mgh, which was converted by his starting kinetic energy, 1/2 mv^2. Set those equal to each other and you can cancel out mass. So you have gh = 1/2 v^2. This avoids having to calculate Time as you did in your approach. Just always remember the relationships of kinetic and potential energies and you can often save time.

oh wow what a great way to look at the question thank you!