Answer:
The range is maximum when the angle of projection is 45 degree.
Explanation:
The formula for the horizontal range of the projectile is given by
The range should be maximum if the value of Sin2θ is maximum.
The maximum value of Sin2θ is 1.
It means 2θ = 90
θ = 45
Thus, the range is maximum when the angle of projection is 45 degree.
If the angle of projection is 0 degree
R = 0
It means the horizontal distance covered by the projectile is zero, it can move in vertical direction.
If the angle of projection is 30 degree.
R = 0.088u^2
If the angle of projection is 45 degree.
R = u^2 / g
Answer:
a) 2.5m/s
b) 0.91m/s
c) 0m/s
Explanation:
Average velocity can be said to be the ratio of the displacement with respect to time.
Average speed on the other hand is the ratio of distance in relation to time
Thus, to get the average velocity for the first half of the swim
V(average) = displacement of first trip/time taken on the trip
V(average) = 50/20
V(average) = 2.5m/s
Average velocity for the second half of the swim will be calculated in like manner, thus,
V(average) = 50/55
V(average) = 0.91m/s
Average velocity for the round trip will then be
V(average) = 0/75, [50+25]
V(average) = 0m/s
Answer:
22.7 meters
Explanation:
Let's remind the difference between distance and displacement:
- distance: the total distance travelled by an object in all its paths
- displacement: the different between the final and initial position of the object
In this case, the problem asks to find the distance covered by the ball. This will be the sum of the distances covered by the ball in each part of its motion, therefore:
(instead, the displacement will be the difference between the final and initial position of the ball, therefore:
)
1). both
2). Venus
3). Venus catastrophically; Earth too but much less.
4). Earth
5). Earth
6). Venus (It would be pretty hard for US to mistake Earth for a star.)
-3 m/s
---------
per min
oh I think 8m/s to 3m/s to 0m/s
idk probably -0.08
