Explanation:
Given that,
A ball is tossed straight up with an initial speed of 30 m/s
We need to find the height it will go and the time it takes in the air.
At its maximum height, its final speed, v = 0 and it will move under the action of gravity. Using equation of motion :
v = u +at
Here, a = -g
v = u -gt
i.e. u = gt
So, the time for upward motion is 3.06 seconds. It means that it will in air for 3.06×2 = 6.12 seconds
Let d is the maximum distance covered by it.
Putting all values
Hence, it will go to a height of 45.91 m and it will in the air for 6.12 seconds.
Explanation:
Starting position at x = 16m
Ending position at x = -25m
Time of flight = 4s
Unknown:
Distance flown = ?
Displacement = ?
Speed = ?
Velocity = ?
Solution:
To find the distance flown, we should understand that the body is moving on the x - plane;
So distance = 16 + 25 = 41m
Displacement is 41m to the left or -x axis
Speed is the distance divided by the time taken;
Speed = = = 10.25m/s
Velocity is 10.25m/s along -x axis
Answer:
Y component = 32.37
Explanation:
Given:
Angle of projection of the rocket is,
Initial velocity of the rocket is,
A vector at an angle with the horizontal can be resolved into mutually perpendicular components; one along the horizontal direction and the other along the vertical direction.
If a vector 'A' makes angle with the horizontal, then the horizontal and vertical components are given as:
Here, as the velocity is a vector quantity and makes an angle of 33.6 with the horizontal, its Y component is given as:
Plug in the given values and solve for . This gives,
Therefore, the Y component of initial velocity is 32.37.
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Answer:
the bike would not be moving
Explanation: