Max height of a projectile?

Question

At the Earth's surface a projectile is launched straight up at a speed of 9.6 km/s. To what height will it rise? Ignore air resistance and the rotation of the Earth.

Answer 1

For _slower_ speeds, you could use this standard formula:

h = v²/(2g)

But that assumes that the gravity is uniform (i.e. that "g" is the same for the entire trajectory). This won't work in this case, because the speed is so great that the projectile will rise to a height where the value of "g" changes significantly. So in that case, it's useful to use the conservation of energy principle:

ΔKE = ΔPE

mv²/2 = integral(F·dr)

Where F = GMm/r²

You can work out the details yourself; but in the end you get the following equation for the height:

h = Rv² / (2gR − v²)

Where:
h = max height
R = earth's radius
v = initial speed
g = acceleration due to gravity at earth's surface.

Answer 2

continually the utmost top is attained while the horizontal displacement is 0.5 the selection. you need to use this fact on your calculations. The time of flight T (the time for the projectile to realize the preliminary horizontal airplane ) is T = 2 U sin ? /g The horizontal selection R = (U^2 /g )*sin 2 ? the utmost top H = (U sin ?) ^2 / (2g) you could combine those 3 right into a simplified way as R tan ? = 4 H = g T^2 / 2 ========================= on the grounds which you opt for a relation between H and R it somewhat is barely tan ? = 4H / R 4H/R could be written as = 2 H / ( R/2) to that end the ratio of two cases the utmost top and nil.5 the selection supplies tan ?

Answer 3

V^2 = (Vo)^2 - 2 * g * x

V = final velocity
Vo = initial Velocity
g = acceleration due to gravity
x = vertical distance (height)

Key point - at the maximum height the velocity is zero.

0 = (Vo)^2 - 2 * g * x
x = ((Vo)^2) / (2 * g)
= (9.6(10^3))^2 / (2 * 9.807)

Xmax = 4,698.7 km (assuming no air resistance, and non-varying gravity field)