Answer:
32.3 m/s
Explanation:
The ball follows a projectile motion, where:
- The horizontal motion is a uniform motion at costant speed
- The vertical motion is a free fall motion (constant acceleration)
We start by analyzing the horizontal motion. The ball travels horizontally at constant speed of
and it covers a distance of
d = 165 m
So, the total time of flight of the ball is
In order to find the vertical velocity of the ball, we have now to analyze its vertical motion.
The vertical motion is a free-fall motion, so the ball is falling at constant acceleration; therefore we can use the following suvat equation:
where
is the vertical velocity at time t
is the initial vertical velocity
is the acceleration of gravity (taking downward as positive direction)
Substituting t = 3.3 s (the time of flight), we find the final vertical velocity of the ball:
It is either the mass or density, I believe it is 3 density tho but it could be mass good luck:)
As per the question a frog jumps 5 m towards east.
Frog again jumps 2 m north.
Let the displacement along east is denoted by vector A and the displacement towards north is denoted as vector B.
Hence magnitude of A = 5 m
Magnitude of B = 2 m
We are asked to calculate the total displacement.
Here the angle between them is 90 degree as A is towards east and B is towards north.
As per parallelogram law of vector addition,the magnitude of total displacement [R] will be-
[cos90= 0]
[ans]
Answer:
The maximum speed is 21.39 m/s.
Explanation:
Given;
radius of the flat curve, r₁ = 150 m
maximum speed, 
= 32.5 m/s
The maximum acceleration on the unbanked curve is calculated as;
the radius of the second flat curve, r₂ = 65.0 m
the maximum speed this unbanked curve should be rated is calculated as;
Therefore, the maximum speed is 21.39 m/s.
