# Projectile motion where yi doesn't equal yf

## Homework Statement

A hammer bro is throwing hammers with a velocity of 1.7 E 1 m/s at 51 degrees down at Mario from a ledge that is 8.5 m tall. Assume that gravitational acceleration in the Mushroom Kingdom is the same as on Earth.
Determine how long each hammer in in the air.

## Homework Equations

vfy^2=viy^2+2g ⋅Δ(y)
vfy=gt+viy

## The Attempt at a Solution

I tried the method of using the equation: vfy^2=viy^2+2g⋅Δ(y) to find vfy; then I substituted the known values so that it would be vfy^2=(17⋅sin51)^2⋅2(-9.8)(8.5) which I found that vfy=2.818375277.
Then I substituted this value to the equation vfy=gt+viy which gave me 2.818375277=-9.8t+(17⋅sin51); this gave me the time of 1.06052127. However, when I input this solution into the answer option it says that I was incorrect. I don't know what/how I messed up, help would be greatly appreciated.

PeroK
Homework Helper
Gold Member
2021 Award

## Homework Statement

A hammer bro is throwing hammers with a velocity of 1.7 E 1 m/s at 51 degrees down at Mario from a ledge that is 8.5 m tall. Assume that gravitational acceleration in the Mushroom Kingdom is the same as on Earth.
Determine how long each hammer in in the air.

## Homework Equations

vfy^2=viy^2+2g ⋅Δ(y)
vfy=gt+viy

## The Attempt at a Solution

I tried the method of using the equation: vfy^2=viy^2+2g⋅Δ(y) to find vfy; then I substituted the known values so that it would be vfy^2=(17⋅sin51)^2⋅2(-9.8)(8.5) which I found that vfy=2.818375277.
Then I substituted this value to the equation vfy=gt+viy which gave me 2.818375277=-9.8t+(17⋅sin51); this gave me the time of 1.06052127. However, when I input this solution into the answer option it says that I was incorrect. I don't know what/how I messed up, help would be greatly appreciated.

You have ##g = -9.8 m/s^2##, which means "down" is the negative direction. Have you been consistent in this?

Note that units are important.

• OnePunch
I believe so as in the first equation I used: vfy^2= (17⋅sin51)^2⋅-19.6⋅8.5. Is is acceptable for me to have 2g as -19.6, or is that where I am faltering? The solution for vfy=2.818375277; and in the second equation I used: vfy=gt+viy→vfy-viy=gt→(vfy-viy)/g=t. This gives me -10.39310607/-9.8=t, then t= 1.06052127.

PeroK
Homework Helper
Gold Member
2021 Award
I believe so as in the first equation I used: vfy^2= (17⋅sin51)^2⋅-19.6⋅8.5. Is is acceptable for me to have 2g as -19.6, or is that where I am faltering? The solution for vfy=2.818375277; and in the second equation I used: vfy=gt+viy→vfy-viy=gt→(vfy-viy)/g=t. This gives me -10.39310607/-9.8=t, then t= 1.06052127.

What have you got for the displacement in the y-direction?

By the way, when I asked whether you had been consistent, that was just a polite way of saying that you hadn't been consistent!

• OnePunch
Should Δy be -8.5 instead of regular 8.5? And I don't see where I messed up with the -9.8m/s^2 consistency.

PeroK
Homework Helper
Gold Member
2021 Award
Should Δy be -8.5 instead of regular 8.5? And I don't see where I messed up with the -9.8m/s^2 consistency.

Well, that's it. If ##g = -9.8 m/s^2##, then ##\Delta y = -8.5m##, as they are both downward.

You need to be careful with the signs of your initial and final velocities as well. For example:

## Homework Statement

I found that vfy=2.818375277.

You have a positive value there, but ##v_{fy}## must be negative.

• OnePunch