Cl2 is the answer. Hope this helps you.
Vo = 5.89 m/s Y = 1.27 m g = 9.81 m/s^2
Time to height
Tr = Vo / g Tr = (5.89 m/s) / (9.81 m/s^2) Tr = 0.60 s
Max height achieved is:
H = Vo^2 / [2g] H = (5.89 )^2 / [ 2 * (9.81) ] H = (34.69) / [19.62] H = 1.77 m
It falls that distance, minus Andrew's catch distance:
h = H - Y h = (1.77 m) - (1.27 m) h = 0.5 m
Time to descend is therefore:
Tf = √ { [2h] / g ] Tf = √ { [ 2 * (0.5 m) ] / (9.81 m/s^2) } Tf = √ { [ 1.0 m ] / (9.81 m/s^2) } Tf = √ { 0.102 s^2 } Tf = 0.32 s
Total time is rise plus fall therefore:
Tt = Tr + Tf Tt = (0.60 s) + (0.32 s) Tt = 0.92 s (ANSWER)
Purpose;
They tend to damp the oscillations(bumpiness) that a car does when it goes over a bump
You can tell that they are worn out by...feeling the car's bumpiness when it goes over a bump...in which the car tend to keep oscillating for a longer period of time
Answer:
(a) the average acceleration of the bus while braking is 8.333 m/s
(b) if the bus took twice as long to stop, the acceleration will be half of the value obtained in part a. [¹/₂ (8.333 m/s) = 4.16 s]
Explanation:
Given;
initial velocity of the bus, v = 25 m/s
time of the motion, t = 3 s
(a) the average acceleration of the bus while braking
a = dv/dt
where;
a is the bus acceleration
dv is change in velocity
dt is change in time
a = 25 / 3
a = 8.333 m/s
(b) If the bus took twice as long to stop, the duration = 2 x 3s
a = 25 / (2 x 3s)
a = ¹/₂ x (25 / 3)
a = ¹/₂ (8.333 m/s) = 4.16 s
Thus, if the bus took twice as long to stop, the acceleration will be half of the value obtained in part a.
