a)Find the time until projectile hits the ground

Use Y components

\(\displaystyle{Y}_{{f}}={Y}_{{i}}+{V}_{{i}}{T}+{\frac{{{1}}}{{{2}}}}{g}{T}^{{2}}\)

\(\displaystyle{Y}_{{f}}=-{125}{m}\)

\(\displaystyle{Y}_{{i}}={0}\)

\(\displaystyle{V}_{{i}}={65.0}{\sin{{37}}}\) make sure your calculator is set to degrees

\(\displaystyle{g}=-{9.8}\ \frac{{m}}{{s}^{{2}}}\)

Solve for T

b)To find range use X coordinates

\(\displaystyle{X}_{{f}}={X}_{{i}}+{V}_{{i}}{T}+{\frac{{{1}}}{{{2}}}}{a}{T}^{{2}}\)

\(\displaystyle{X}_{{i}}={0}\)

\(\displaystyle{V}_{{i}}={65.0}{\cos{{37}}}\)

a=0 this is always the case in projectile motion

Plug T from a) and find \(\displaystyle{X}_{{f}}\)

c)Remember there is no air friction on this problem, so accelaration is zero. That means velocity stays the same on the X coordinate.

As for the Y coordinate \(\displaystyle{V}_{{f}}={V}_{{i}}+{a}{T}\)

You know T, a=-g, and \(\displaystyle{V}_{{i}}\)

The direction is down or course.

d)The magnitude is found by \(\displaystyle\sqrt{{{y}^{{2}}+{x}^{{2}}}}\) you know the drill

e) I think this is \(\displaystyle\theta={\arccos{}}\) (x/magnitude found in d)

f) \(\displaystyle{V}_{{f}}={V}_{{i}}+{a}{t}\) use Y components

\(\displaystyle{V}_{{f}}={0}\) at the top

we know \(\displaystyle{V}_{{i}}\) and a=-g

find this T and use in the position formula to get \(\displaystyle{Y}_{{f}}\)

Use Y components

\(\displaystyle{Y}_{{f}}={Y}_{{i}}+{V}_{{i}}{T}+{\frac{{{1}}}{{{2}}}}{g}{T}^{{2}}\)

\(\displaystyle{Y}_{{f}}=-{125}{m}\)

\(\displaystyle{Y}_{{i}}={0}\)

\(\displaystyle{V}_{{i}}={65.0}{\sin{{37}}}\) make sure your calculator is set to degrees

\(\displaystyle{g}=-{9.8}\ \frac{{m}}{{s}^{{2}}}\)

Solve for T

b)To find range use X coordinates

\(\displaystyle{X}_{{f}}={X}_{{i}}+{V}_{{i}}{T}+{\frac{{{1}}}{{{2}}}}{a}{T}^{{2}}\)

\(\displaystyle{X}_{{i}}={0}\)

\(\displaystyle{V}_{{i}}={65.0}{\cos{{37}}}\)

a=0 this is always the case in projectile motion

Plug T from a) and find \(\displaystyle{X}_{{f}}\)

c)Remember there is no air friction on this problem, so accelaration is zero. That means velocity stays the same on the X coordinate.

As for the Y coordinate \(\displaystyle{V}_{{f}}={V}_{{i}}+{a}{T}\)

You know T, a=-g, and \(\displaystyle{V}_{{i}}\)

The direction is down or course.

d)The magnitude is found by \(\displaystyle\sqrt{{{y}^{{2}}+{x}^{{2}}}}\) you know the drill

e) I think this is \(\displaystyle\theta={\arccos{}}\) (x/magnitude found in d)

f) \(\displaystyle{V}_{{f}}={V}_{{i}}+{a}{t}\) use Y components

\(\displaystyle{V}_{{f}}={0}\) at the top

we know \(\displaystyle{V}_{{i}}\) and a=-g

find this T and use in the position formula to get \(\displaystyle{Y}_{{f}}\)