The time of motion of the object at the given speed and distance is 4 seconds.
<h3>Time of motion of the object</h3>
The time of motion of the object is calculated as follows;
t = d/v
where;
- d is distance = 10 m
- v is speed = 2.5 m/s
- t is time of motion
t = 10/2.5
t = 4 s
Thus, the time of motion of the object at the given speed and distance is 4 seconds.
Learn more about time here: brainly.com/question/2854969
#SPJ1
Answer:
the cannonball’s velocity parallel to the ground is 86.6m/S
Explanation:
Hello! To solve this problem remember that in a parabolic movement the horizontal component X of the velocity of the cannonball is constant while the vertical one varies with constant acceleration.
For this case we must draw the velocity triangle and find the component in X(see atached image).
V= Initial velocity=100M/S
V= Initial velocity=100M/S
Vx=cannonball’s velocity parallel to the ground
Solving for Vx
Vx=Vcos30
Vx=(100m/S)(cos30)=86.6m/s
the cannonball’s velocity parallel to the ground is 86.6m/S
Water - helps with dehydration.
instruction - help you understand how you need to do the workout.
trainers - wether it's an online trainer or a gym trainer they always say how to be safe while working out.
Answer:
0.303s
Explanation:
horizontal distance travel = 2.050 m, vertical distance travel = 0.45 m
Using equation of linear motion
Sy = Uy t + 1/2 gt² Uy is the inital vertical component of the velocity, t is the time taken for the vertical motion in seconds, and S is the vertical distance traveled, taken downward vertical motion as negative
-0.45 = 0 - 0.5 × 9.81×t²
0.45 / (0.5 × 9.81) = t²
t = √0.0917 = 0.303 s
3 as a single number is considered a sf
