Displacement, velocity and acceleration

Question

A large boulder is ejected vertically upward from a volcano with an initial speed of 50 m/s. Ignoring air resistance:

a) At what time after being ejected is the boulder moving at 25 m/s upward?

b) At what time is it moving at 25 m/s downward?

c) When is the displacement of the boulder from its initial position zero?

d) When is the velocity of the boulder zero?

What are the magnitude and direction of the acceleration while the boulder is:

i. Moving upward (9.8 m/s² downward)

ii. Moving downward (9.8 m/s² downward)

iii. At the highest point (9.8 m/s² downward)

Answer 1

v = -g.t + v₀ on (Oy) ascendant axis with g=9.8 m/s² and v₀=50 m/s (in accordance with F=m.a and F=P=m.g)

a - v = +v₀/2 = 25 m/s so v₀/2 = -g.t + v₀ and t = v₀/(2.g) = 2.55 s = 2.6 s.

b - v = -v0/2 = -25 m/s so -v₀/2 = -g.t + v₀ and t = 3v₀/(2g) = 7.65 s = 7.7 s.

c - z = -(1/2).g.t² + v₀.t = 0 so t.(-(1/2).g.t + v₀) = 0 so t = 0 s or t = 2.v₀/g = 10.2 s.

d - v = 0 = -g.t +v₀ so t = v₀/g = 5.1 s.

i - On (Oy) ascendant axis : g = -9.8 m/s² (magnitude is g = 9.8 m/s2 and direction is downward).

ii - On (Oy) ascendant axis : g = -9.8 m/s² (magnitude is g = 9.8 m/s2 and direction is downward) (same : acceleration doesn't change when the boulder is going upward or downward because the strength is still P=m.g with same downward direction so that a=g because F=m.a=P=m.g or a = -g on (Oy)).

iii - On (Oy) ascendant axis : g = -9.8 m/s² (magnitude is g = 9.8 m/s2 and direction is downward).