Answer:
a)
Y0 = 0 m
Vy0 = 15 m/s
ay = -9.81 m/s^2
b) 7.71 m
c) 3.06 s
Explanation:
The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards
Y(0) = 0 m
Vy(0) = 15 m/s
ay = -9.81 m/s^2 (negative because it points down)
Since acceleration is constant we can use the equation for uniformly accelerated movement:
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
To find the highest point we do the first time derivative (this is the speed:
V(t) = Vy0 + a * t
We equate this to zero
0 = Vy0 + a * t
0 = 15 - 9.81 * t
15 = 9.81 * t
t = 0.654 s
At this time it will have a height of:
Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m
The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.
0 = Y0 + Vy0 * t + 1/2 * a * t^2
0 = 0 + 15 * t - 1/2 * 9.81 t^2
0 = 15 * t - 4.9 * t^2
0 = t * (15 - 4.9 * t)
t1 = 0 This is the moment it jumped into the air
0 = 15 - 4.9 * t2
15 = 4.9 * t2
t2 = 3.06 s This is the moment when it falls again.
3.06 - 0 = 3.06 s
The first Olympic Games were in Greece
Answer:
d A ball is rolling down an inclined plane.
Explanation:
When path length is equal to the displacement
then we can say that the motion of the object must be in straight line so that the distance and displacement must be same
SO here we can say
a A ball on the end of a string is moving in a vertical circle.
In circular path distance and displacement is not same
b A toy train is traveling around a circular track.
In circular path distance and displacement is not same
c A train travels 5 miles east before it stops. It then travels 2 miles west.
Net displacement is 3 miles East while distance is 7 miles
d A ball is rolling down an inclined plane.
Here its motion is in straight line so we can say that path length and displacement will be same
e A ball rises and falls after being thrown straight up from the earth's surface.
In this type of to and fro motion path length is not same as displacement
Answer:
Explanation:
initial angular velocity, ωo = 0 rad/s
angular acceleration, α = 30.5 rad/s²
time, t = 9 s
radius, r = 0.120 m
let the velocity is v after time 9 s.
Use first equation of motion for rotational motion
ω = ωo + αt
ω = 0 + 30.5 x 9
ω = 274.5 rad/s
v = rω
v = 0.120 x 274.5
v = 32.94 m/s
