Step 1

Givens:

\(\displaystyle\theta_{{{0}}}={53.0}^{{{0}}}\)

\(\displaystyle{h}={730}{m}\)

\(\displaystyle{t}={5.00}{s}\)

Step 2

Part a:

The speed of the plane is the initial speed of the projectile which can be obtained from:

\(\displaystyle{y}-{y}_{{{0}}}={v}_{{{0}{y}}}{t}-{\frac{{{1}}}{{{2}}}}{gt}^{{{2}}}\)

Where \(\displaystyle{y}={0},\ {y}_{{{0}}}={h}\), therefore \(\displaystyle{v}_{{{0}{y}}}\) equals:

\(\displaystyle{v}_{{{0}{y}}}={\frac{{{\frac{{{1}}}{{{2}}}}{gt}^{{{2}}}-{h}}}{{{t}}}}={\frac{{{\left({0.5}\right)}{\left({9.8}\right)}{\left({5}\right)}^{{{2}}}-{730}}}{{{5}}}}=-{121.5}\frac{{m}}{{s}}\)

The angle with the horizontal is \(\displaystyle{90}-{53}={37}\)

And since it is downward it is going to be \(\displaystyle\theta=-{37}^{{{0}}}\)

So now we have the y-component of the velocity and the angle this means we can find the total speed from the relation

\(\displaystyle{v}_{{{0}}}={\frac{{{v}_{{{0}{y}}}}}{{{\sin{{\left(-{37}\right)}}}}}}={\frac{{-{121.5}\frac{{m}}{{s}}}}{{{\sin{{\left(-{37}\right)}}}}}}={201}\frac{{m}}{{s}}\)

Step 3

Part b:

The horizontal distance it covers is:

\(\displaystyle{x}={x}_{{{0}}}+{v}_{{\otimes}}{t}\)

Where \(\displaystyle{x}_{{{0}}}={0}\) therefore:

\(\displaystyle{x}={v}_{{{0}}}{\cos{\theta}}{t}={\left({201}\frac{{m}}{{s}}\right)}{\left({\cos{-}}{37}\right)}{\left({5}{s}\right)}={802}{m}\)

Step 4

Part c:

The horizontal component of the velocity doesn't change, so it is given by:

\(\displaystyle{v}_{{{x}}}={v}_{{{0}{x}}}={\left({201}\frac{{m}}{{s}}\right)}{\left({\cos{-}}{37}\right)}={160}\frac{{m}}{{s}}\)

Step 5

Part d:

Meanwhile the vertical comonent is given by:

\(v_{y}=v_{0y}-gt=-121.5m/s-(9.8m/s^{2})(5s)=-170m/s\)