A tourist drops a rock from rest from a guardrail at an observatory point overlooking a valley. Calculate the velocity of the rock at the following times: a) 2.0 s b) 6.0 s 6. Determine the displacement for the rock in question 6 at

a) 19.6 m/s, 19.6 m

b) 58.8 m/s, 176.4 m

Explanation:

Take down to be positive.  Given:

v₀ = 0 m/s

a = 9.8 m/s²

a) Find v and Δy when t = 2.0 s.

v = at + v₀

v = (9.8 m/s²) (2.0 s) + 0 m/s

v = 19.6 m/s

Δy = v₀ t + ½ at²

Δy = (0 m/s) (2.0 s) + ½ (9.8 m/s²) (2.0 s)²

Δy = 19.6 m

b) Find v and Δy when t = 6.0 s.

v = at + v₀

v = (9.8 m/s²) (6.0 s) + 0 m/s

v = 58.8 m/s

Δy = v₀ t + ½ at²

Δy = (0 m/s) (6.0 s) + ½ (9.8 m/s²) (6.0 s)²

Δy = 176.4 m

i dont know but

explanation:

whoever your teacher is, deserves a raise for using pokemon as an example

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s