Answer:
∅ = 89.44°
Explanation:
In situations like this air resistance are usually been neglected thereby making g= 9.81 m/[tex]s^{2}[/tex]
Bring out the given parameters from the question:
Initial Velocity ([tex]V_{1}[/tex]) = 1000 m/s
Target distance (d) = 2000 m
Target height (h) = 800 m
Projection angle ∅ = ?
Horizontal distance = [tex]V_{1x}[/tex]tcos ∅ .......................... Equation 1
where [tex]V_{1x}[/tex] = velocity in the X - direction
t = Time taken
Vertical Distance = y = [tex]V_{1y}[/tex] t - [tex]\frac{1}{2}[/tex]g[tex]t^{2}[/tex] ................... Equation 2
Where [tex]V_{1y}[/tex] = Velocity in the Y- direction
t = Time taken
[tex]V_{1y}[/tex] = [tex]V_{1}[/tex]sin∅
Making time (t) subject of the formula in Equation 1
t = d/([tex]V_{1x}[/tex]cos ∅)
t = [tex]\frac{2000}{1000coso}[/tex] = [tex]\frac{2}{cos0}[/tex] = [tex]\frac{d}{cos o}[/tex] ...................Equation 3
substituting equation 3 into equation 2
Vertical Distance = d = [tex]V_{1y}[/tex] [tex]\frac{d}{cos o}[/tex] - [tex]\frac{1}{2}[/tex]g[tex]\frac{2}{cos0} ^{2}[/tex]
Vertical Distance = h = sin∅ [tex]\frac{d}{cos o}[/tex] - [tex]\frac{1}{2}[/tex]g[tex]\frac{2}{cos0} ^{2}[/tex]
Vertical Distance = h = dtan∅ - [tex]\frac{1}{2}[/tex]g[tex]\frac{2}{cos0} ^{2}[/tex]
Applying geometry
[tex]\frac{1}{cos o}[/tex] = [tex]tan^{2} o[/tex] + 1
Vertical Distance = h = d tan∅ - 2 g ([tex]tan^{2} o[/tex] + 1)
substituting the given parameters
800 = 2000 tan ∅ - 2 (9.81)( [tex]tan^{2} o[/tex] + 1)
800 = 2000 tan ∅ - 19.6( [tex]tan^{2} o[/tex] + 1) Equation 4
Replacing tan ∅ = Q .....................Equation 5
In order to get a quadratic equation that can be easily solve.
800 = 2000 Q - 19.6[tex]Q^{2}[/tex] + 19.6
Rearranging 19.6[tex]Q^{2}[/tex] - 2000 Q + 780.4 = 0
[tex]Q_{1}[/tex] = 101.6291
[tex]Q_{2}[/tex] = 0.411
Inserting the value of Q Into Equation 5
tan ∅ = 101.63 or tan ∅ = 0.4114
Taking the Tan inverse of each value of Q
∅ = 89.44° ∅ = 22.37°
