Given:
(Initial velocity)u=20 m/s
At the maximum height the final velocity of the ball is 0.
Also since it is a free falling object the acceleration acting on the ball is due to gravity g.
Thus a=- 9.8 m/s^2
Now consider the equation
v^2-u^2= 2as
Where v is the final velocity which is measured in m/s
Where u is the initial velocity which is measured in m/s
a is the acceleration due to gravity measured in m/s^2
s is the displacement of the ball in this case it is the maximum height attained by the ball which is measured in m.
Substituting the given values in the above formula we get
0-(20x20)= 2 x- 9.8 x s
s= 400/19.6= 20.41m
Thus the maximum height attained is 20.41 m by the ball
Answer:
Your answer is D It does not need to be repeatable.
the reason for this is because C is correct.
You need to be able to experiment on something multiple times so that you can gather further data to imbedded your evidence in facts.
Answer:
A. 50 m/s
Explanation:
Given in the y direction:
v₀ = 0 m/s
a = 10 m/s²
t = 4 s
Find: v
v = at + v₀
v = (10 m/s²) (4 s) + 0 m/s
v = 40 m/s
In the x direction, the velocity is constant at 30 m/s.
The overall speed is:
v² = (30 m/s)² + (40 m/s)²
v = 50 m/s
Answer:
Angular acceleration, is 
Explanation:
Given that,
Initial speed of the drill, 
After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min, final angular speed, 
We need to find the drill’s angular acceleration. It is given by the rate of change of angular velocity.

So, the drill's angular acceleration is
.
Answer:
1.08 m/s
Explanation:
This can be solved with two steps, first we need to find the time taken to fall 9.5 m, then we can divide the horizontal distance covered with time taken to calculate the velocity.
Time taken to fall 9.5 m
vertical acceleration = a = 9.8 m/s^2.
vertical velocity = 0, (since there is only horizontal component for velocity,
)
distance traveled s = 9.5 m.
Substituting these values in the equation



⇒ t= 1.392 sec
Velocity needed
We know the time taken (1.392 s) to travel 1.5 m,
So velocity = 1.5 m / 1.392 s = 1.08 m/s
hence velocity of the diver must be at least 1.08 m/s