# A plane, diving with constant speed at an angle of 53.0^{\circ}

A plane, diving with constant speed at an angle of $$\displaystyle{53.0}^{{\circ}}$$ with the vertical, releases a projectile at an altitude of 730 m. The projectile hits the ground 5.00 s after release.
a) What is the speed of the plane?
b) How far does the projectile travel horizontally during its flight? What are the
c) horizontal and (d) vertical components of its velocity just before striking the ground?

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Lauren Fuller

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

###### Not exactly what you’re looking for?
Camem1937
Step 1
In order to answer the first question we need to calculate the initial speed of the projectile because it's the same speed of the plane.
$$\displaystyle{V}_{{{f}{y}^{{{2}}}}}={V}_{{{o}{y}^{{{2}}}}}+{2}\times{a}\times{d}$$
$$\displaystyle{V}_{{{f}{y}}}={0}$$
$$\displaystyle{V}_{{{o}{y}}}=\sqrt{{-{2}\times{\left(-{9.8}\right)}\times{730}}}$$
$$\displaystyle{V}_{{{o}{y}}}={119.61}\frac{{m}}{{s}}$$
Step 2
The velocity has to be negative because is going down so, $$\displaystyle{V}_{{{o}{y}}}=-{119.61}\frac{{m}}{{s}}$$ now that we have the Y component of velocity we can get the plane velocity by:
the angle with the horizontal is $$\displaystyle\propto=-{90}+{53}=-{37}$$
$$\displaystyle{V}_{{{o}{y}}}={V}\times{\cos{{\left(\alpha\right)}}}$$
$$\displaystyle{V}={\frac{{{V}_{{{o}{y}}}}}{{{\sin{{\left({53}\right)}}}}}}$$
$$\displaystyle{V}={198.75}\frac{{m}}{{s}}$$
Now that we have the Initial velocity we can calculate the horizontal displacement:
$$\displaystyle{d}={V}\times{\cos{{\left(\alpha\right)}}}\times{t}$$
$$\displaystyle{d}={198.75}\times{\cos{{\left(-{37}\right)}}}\times{5}$$
$$\displaystyle{d}={793.7}{m}{t}{s}$$
Because the horizontal velocity remains the same, it is given by:
$$\displaystyle{V}_{{{x}}}={V}\times{\cos{{\left(-{37}\right)}}}={158.72}\frac{{m}}{{s}}$$
In order the calculate the Y componet we will use the next formula:
$$\displaystyle{V}_{{{f}{y}}}={V}_{{{o}}}+{a}\times{t}$$
$$\displaystyle{V}_{{{f}{y}}}=-{119.71}+{\left(-{9.8}\right)}\times{5}$$
$$\displaystyle{V}_{{{f}{y}}}=-{168.71}\frac{{m}}{{s}}$$