I failure to find an observation related to motion in two dimension projectile motion.
Prove That/Show that
maximum range =4Height
or
Rmax=4H
please help me thanks Ahsan Khan
Professor Hu did it here. Take his derivation for max range, and height, and show that Rmax=4H
www-hep.uta.edu/~yu/teaching/summer08-1441-001/lectures/phys1441-summer08-060308.ppt -
That link did not work
vertical v = Vo - gt
at the top, H, v = 0
so
t = Vo/g
then H = Vo t - .5 g t^2
H = Vo^2/g -.5 Vo^2/g = .5 Vo^2/g
that t is time to the top, H, which is half the total time in the air
u = Uo constant horizontal speed
R = range = Uo *2t = 2 Uo t = 2 UoVo/g
so
H/R = .5Vo/2 Uo = (1/4) Vo/Uo
I suspect that you class has already done the angle for maximum range is 45 degrees above horizontal where Vo = Uo
so
H/R = 1/4
Anyway in case you have not done it, for speed S and elevation angle T:
Uo = s cos T
Vo = s sin T
for max height
t = (s/g) sin T like before
R = 2 Uo t = 2 s cos T (s/g )sin T
for max
dR/dT = 0
(2 s^2/g)(-sin^2 T + cos^2 T)
or max range when T = 45 degrees and
Uo = Vo
i don't know the answer but how 4 the man height varies
To prove that the maximum range of a projectile in two-dimensional motion is equal to 4 times the height (Rmax = 4H), we can use the equations of projectile motion.
Projectile motion consists of two independent motions: horizontal motion with constant velocity and vertical motion with constant acceleration due to gravity.
Let's assume the following variables:
Initial velocity = v
Launch angle = θ
Height = H
Maximum range = Rmax
First, we need to find the time taken for the projectile to reach its maximum height (t1). To do this, we'll use the equation for vertical velocity:
v_vertical = v * sin(θ)
At the maximum height, the vertical velocity becomes zero. Therefore:
0 = v * sin(θ) - g * t1
Rearranging this equation, we have:
t1 = v * sin(θ) / g
Next, we'll find the time of flight (t_total), which is the total time the projectile is in the air. We can calculate this using the equation for time of flight:
t_total = 2 * t1
Substituting the value of t1, we have:
t_total = 2 * (v * sin(θ) / g)
Now, we can find the horizontal distance traveled by the projectile, which is the range (Rmax). Using the equation for horizontal distance:
Rmax = v * cos(θ) * t_total
Substituting the value of t_total, we get:
Rmax = v * cos(θ) * 2 * (v * sin(θ) / g)
Simplifying this expression:
Rmax = (2 * v^2 * sin(θ) * cos(θ)) / g
Since sin(2θ) = 2 * sin(θ) * cos(θ), we can rewrite the equation as:
Rmax = (v^2 * sin(2θ)) / g
Using the trigonometric identity sin(2θ) = 2 * sin(θ) * cos(θ), we can further simplify the equation:
Rmax = (v^2 * 2 * sin(θ) * cos(θ)) / g
Finally, using the identity sin(θ) = H / v and cos(θ) = sqrt(1 - sin^2(θ)), we can substitute these values in the equation:
Rmax = (v^2 * 2 * (H / v) * sqrt(1 - (H / v)^2)) / g
Simplifying further:
Rmax = (2 * H * sqrt(v^2 - H^2)) / g
Given that Rmax = 4H, we can equate both equations:
4H = (2 * H * sqrt(v^2 - H^2)) / g
2 * g * H = 2 * H * sqrt(v^2 - H^2)
Simplifying further:
g = sqrt(v^2 - H^2)
Squaring both sides:
g^2 = v^2 - H^2
Rearranging this equation, we have:
v^2 = g^2 + H^2
This equation represents the square of the initial velocity of the projectile. Since the velocity cannot be negative, we take the positive square root:
v = sqrt(g^2 + H^2)
Substituting this value back into the equation for Rmax:
Rmax = (2 * H * sqrt(v^2 - H^2)) / g
Rmax = (2 * H * sqrt((sqrt(g^2 + H^2))^2 - H^2)) / g
Rmax = (2 * H * sqrt(g^2 + H^2 - H^2)) / g
Rmax = (2 * H * sqrt(g^2)) / g
Rmax = (2 * H * g) / g
Finally, canceling out the g's, we get:
Rmax = 2H
Thus, Rmax = 2H, not 4H. So it seems there was an error or misunderstanding in the initial statement that Rmax = 4H.