Asked by Emily
I don't understand.
derive a general equation that will predict the horizontal range of a projectile that is fired from one vertical position and ends at any other vertical position with an initial velocity, V1, at an angle, pheta, from the horizontal. The equation should be derived from kinematic equations for projectile motion and should be expressed in terms of v1, pheta, y2, and g.
derive a general equation that will predict the horizontal range of a projectile that is fired from one vertical position and ends at any other vertical position with an initial velocity, V1, at an angle, pheta, from the horizontal. The equation should be derived from kinematic equations for projectile motion and should be expressed in terms of v1, pheta, y2, and g.
Answers
Answered by
Chanz
y =vo(y)t +.5at^2 but y=0 when it lands.
0 = vo(y)t + .5at^2. a is gravity (g) and the solution t = 0 is trivial.
0 = vo(y) +.5gt. But vo in y is just vo*sin(theta)
0= vosin(theta) +.5gt
Solve for t.
t = 2vosin(theta)/g.
Now put this into the equation for x
x = vo(x)t, but vo(x) is vocos(theta) so
x = vocos(theta)*2vosin(theta)/g
there is a trig identity:
2cos(theta)sin(theta) = sin(2theta)
x = vo^2 sin(2theta)/g
Sorry I just realized I used vo instead of v1, but you get the picture.
0 = vo(y)t + .5at^2. a is gravity (g) and the solution t = 0 is trivial.
0 = vo(y) +.5gt. But vo in y is just vo*sin(theta)
0= vosin(theta) +.5gt
Solve for t.
t = 2vosin(theta)/g.
Now put this into the equation for x
x = vo(x)t, but vo(x) is vocos(theta) so
x = vocos(theta)*2vosin(theta)/g
there is a trig identity:
2cos(theta)sin(theta) = sin(2theta)
x = vo^2 sin(2theta)/g
Sorry I just realized I used vo instead of v1, but you get the picture.
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