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:
C. less than 950 N.
Explanation:
Given that
Force in north direction F₁ = 500 N
Force in the northwest F₂ = 450 N
Lets take resultant force R
The angle between force = θ
θ = 45°
The resultant force R


R= 877.89 N
Therefore resultant force is less than 950 N.
C. less than 950 N
Note- When these two force will act in the same direction then the resultant force will be 950 N.
KE = 2000 J
Explanation:
KE = (1/2)mv^2
= (1/2)(0.100 kg)(200 m/s)^2
= 2000 J
Answer:
Paramedic
Explanation:
They will be on the move the whole time
Answer:
The velocity of the truck after this elastic collision is 15.7 m/s
Explanation:
It is given that,
Mass of the car, 
Mass of the truck, 
Initial velocity of the car,
Initial velocity of the truck, u₂ = 0
After the collision the velocity of the car is, v₁ = -11 m/s
Let v₂ is the velocity of the truck after this elastic collision. Using the conservation of momentum as :

So, the velocity of the truck after this elastic collision is 15.7 m/s. Hence, the correct option is (c).