projectile?

Question

If a ball is thrown up with velocity v at an angle thata,
WHY is the smallest kinetic energy .5m(vcosthata)^2 (kinetic energy at the top of trajectory) isnt it .5m(vx)? because the only v would be in the x direction ...... and vx is not vcosthata

also how do you get max height? the answer is (vsinthata^2)/2g

Answer 1

The horizontal component of velocity is
v cos θ and remains the same through out its path.
Vertical component is initially v sin θ and it decreases to zero at a height h from the ground.
Total energy is there fore a constant term of 1/2 m (v cos θ) ^2 and a quantity which reduces to zero at the top.
Naturally this will be the minimum and hence the k.e is the minimum at the top.
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The maximum height is obtained by considering only the vertical motion.
In the vertical direction its initial velocity is v sin θ.
At the maximum height its vertical velocity is zero.
{We don't bother about its horizontal velocity at the top which is equal to v cos θ.}
Acceleration is - g.
Using v^2 - u^2 = 2as
0 - (v sin θ) ^2 = - 2 g h
h = (v sin θ) ^2 / 2g.
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Or if you consider the energy principle:
Initial kinetic energy = 1/2 mv^2.
Final kinetic energy = 1/2 m (v cos θ) ^2

Change in K.E =1/2 mv^2 - 1/2 m (v cos θ) ^2
= 1/2 m v^2 {1- (cos θ) ^2} = 1/2m v*2 (sin θ) ^2.
This is equal to the work done against gravity = mg h.
mg h = 1/2m v*2 (sin θ) ^2
h = (v sin θ) ^2 / 2g
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Answer 2

Assuming that the angle theta is measured upward from the horizontal, Vx is V*cos(theta) and all of your formulas are correct. The ball starts out at angle theta with some vertical velocity and some horizontal velocity. The vertical velocity decays to zero leaving only horizontal velocity at the top of the trajectory. That's why the kinetic energy is lowest at the top. The max height is found by equating the vertical component of the kinetic energy [.5*Mass*{Vsin(theta)} ^2] with the potential energy of increased height [Mass*gravity*height]. Put the equal sign between them, solve for height and you will get the answer (which you already have).

Answer 3

The smallest kinetic energy is at the top of the trajectory because the ball momentarily stops its motion going upward and starts falling downward,max height is when the ball's velocity in the vertical direction is zero ( least kinetic energy)

Answer 4

i just would like to give math relationship leading to Minimum property>>
velocity V(t) at any time is
V^2(t) = vx^2 + vy^2
KE = 0.5m{v cos p]^2 + [v sin p - gt]^2}
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any sum of squares (A^2+B^2) be MINIMUM either when A=0 or B=0
A = not zero >> v cos (p)= 0 means p = 90 fired vertically
B = 0 = v sin p - gt
t = v sin p /g = this is the time taken for reaching the max-height.
KE(min) = 0.5mV^2 (min) = 0.5m v^2 cos^2 p
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max height>>
y = (u sin p) * t - 0.5 gt^2
dy/dt = vy = (u sin p) - gt
at y = h(max), vy =0
>> t = u sin p / g >> this half the time of flight
put in y-equ
h(max) = (u sin p) *(u sin p/g) - 0.5 g[u^2 sin^2/g^2]
h(max) = u^2 sin^2/g) - u^2 sin^2/2g = u^2 sin^2/2g
h(max) = (u sin p)^2 / 2g