Question

projectile problem

Answer 1

The horizontal velocity of the particle will double at the peak.

Answer 2

Ahh conservation of linear momentum for horizontal component ?

Answer 3

maximum height is given by \(h = \dfrac{(u\sin 60)^2}{2g} = \dfrac{(20*\sqrt{3}/2)^2}{2*10} = 15\)

I think this is going to get messy...

Answer 4

how about finding the time to the peak, (when vy=0)
Vy - g t =0
20 sin(60) - 9.8 t =0
t = 10 sqr(3)/9.8 or about sqr(3)

now find the distance horizontally in that time:
Vx * t = 20 cos(60) * sqr(3) = 10 sqr(3)

at the peak, the particle "jumps" to twice the speed, all of it horizontal,
but it continues to fall at the same rate and will take sqr(3) seconds to hit the ground.

distance moved over the second interval is twice as far as the first, i.e. 20 sqr(3)
and total distance is
30 sqr(3)

Answer 5

that looks very nice!