A projectile is fired from the origin with an initial velocity V₁=100m/s at an angle θ₁=30⁰. Another is fired from a point on the x-axis at a distance x₀=60m from the origin with initial velocity V₂=80m/s at an angle of θ₂. It is desired to have the two projectiles collide at some point P(x,y).
A. )What is θ₂?
B.) At what time and point in space do they collide?
C.) Find velocity components of each just before impact.
3 answers
t=3
The projectile fired from the origin impacts 883.67m downrange after reaching a height of 127.55m.
The second projectile will collide with the first projectile before it reaches its maximum height if the second projectile's launch angle was selected to result in equal maximum heights.
Without another boundary condition, there are a multitude of launch angles for the second projectile that will result in an intercept of the first.
Might there be another piece of information that would open the solution window.
The second projectile will collide with the first projectile before it reaches its maximum height if the second projectile's launch angle was selected to result in equal maximum heights.
Without another boundary condition, there are a multitude of launch angles for the second projectile that will result in an intercept of the first.
Might there be another piece of information that would open the solution window.
For instance:
Assume that the second projectile is launched such that it has the same maximum altitude as the first.
We make use of the two equations for height and distance of projectiles:
h = Vo^2(sin(a)^2)/2g and
d = Vo^2(sin(2b))/g
where
h = the maximum height reached – meters
d = the horizontal distance traveled - m
Vo = the initial launch velocity – meters/sec.
a and b = the angle of the velocity vectors to the horizontal
g = the acceleration due to gravity - 9.8m/sec.^2
Given:
Projectile 1………Projectile 2
Vo = 100m/s…...……80m/s
a = 30 deg
b = ……………..….....TBD
g = 9.8m/sec.^2……9.8
100^2(sin30)^2/2(9.8 = 80^2(siny)^2/2(9.8 or y = 38.682 deg.
d1 = 100^2(sin60)/9.8 = 883.67m
d2 = 80^2(sin77.364)/9.8 = 637.24m
Time of intercept
100(cos30)t = 60 + 80(cos938.682)t or t = 2.48 sec.
Point of intercept
x = 100(.866)2.48 = 214.76m
y = 50(2.48) – 4.9(2.48)^2 = 93.86m.
I’ll let you figure out the velocity info.
Note that if the maximum height of the second projectile was allowed to be any quantity, several answers are possible.
Assume that the second projectile is launched such that it has the same maximum altitude as the first.
We make use of the two equations for height and distance of projectiles:
h = Vo^2(sin(a)^2)/2g and
d = Vo^2(sin(2b))/g
where
h = the maximum height reached – meters
d = the horizontal distance traveled - m
Vo = the initial launch velocity – meters/sec.
a and b = the angle of the velocity vectors to the horizontal
g = the acceleration due to gravity - 9.8m/sec.^2
Given:
Projectile 1………Projectile 2
Vo = 100m/s…...……80m/s
a = 30 deg
b = ……………..….....TBD
g = 9.8m/sec.^2……9.8
100^2(sin30)^2/2(9.8 = 80^2(siny)^2/2(9.8 or y = 38.682 deg.
d1 = 100^2(sin60)/9.8 = 883.67m
d2 = 80^2(sin77.364)/9.8 = 637.24m
Time of intercept
100(cos30)t = 60 + 80(cos938.682)t or t = 2.48 sec.
Point of intercept
x = 100(.866)2.48 = 214.76m
y = 50(2.48) – 4.9(2.48)^2 = 93.86m.
I’ll let you figure out the velocity info.
Note that if the maximum height of the second projectile was allowed to be any quantity, several answers are possible.
1 answer