Answer:
y₀ = 10.625 m
Explanation:
For this exercise we will use the kinematic relations, where the upward direction is positive.
y = y₀ + v₀ t - ½ g t²
in the exercise they indicate the initial velocity v₀ = 8 m / s.
when the rock reaches the ground its height is zero
0 = y₀ + v₀ t - ½ g t²
y₀i = -v₀ t + ½ g t²
let's calculate
y₀ = - 8 2.5 + ½ 9.8 2.5²
y₀ = 10.625 m
Answer:im so sorry i cant find anything either ask your teacher for some help is the best thing i can do
Acceleration = Change in Velocity / time
a = (v - u) / t
Where v = final velocity in m/s
u = initial velocity in m/s
t = time in seconds.
a = acceleration in m/s²
A proper record of the changes in velocity with the corresponding time would help find the acceleration.
Answer:
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Answer:
20m/s
Explanation:
acceleration=final velocity-initial velocity/time
4.0m/s²=v m/s-0m/s/5.0sec
5.0sec×4.0m/s²=v m/s-0m/s×5.0m/s/5.0m/s
20m/s=v
