A scientist reports the detection of an atomic particle known as a cosmic ray that came toward his experiment from outer space a
1 answer:
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Answer:
∅ = 89.44°
Explanation:
In situations like this air resistance are usually been neglected thereby making g= 9.81 m/
Bring out the given parameters from the question:
Initial Velocity () = 1000 m/s
Target distance (d) = 2000 m
Target height (h) = 800 m
Projection angle ∅ = ?
Horizontal distance = tcos ∅ .......................... Equation 1
where = velocity in the X - direction
t = Time taken
Vertical Distance = y = t -
g
................... Equation 2
Where = Velocity in the Y- direction
t = Time taken
=
sin∅
Making time (t) subject of the formula in Equation 1
t = d/(cos ∅)
t = =
=
...................Equation 3
substituting equation 3 into equation 2
Vertical Distance = d =
-
g
Vertical Distance = h = sin∅ -
g
Vertical Distance = h = dtan∅ - g
Applying geometry
=
+ 1
Vertical Distance = h = d tan∅ - 2 g ( + 1)
substituting the given parameters
800 = 2000 tan ∅ - 2 (9.81)( + 1)
800 = 2000 tan ∅ - 19.6( + 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 + 19.6
Rearranging 19.6 - 2000 Q + 780.4 = 0
= 101.6291
= 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°