Can you show the diagram please
Answer:
The time after which the two stones meet is tₓ = 4 s
Explanation:
Given data,
The height of the building, h = 200 m
The velocity of the stone thrown from foot of the building, U = 50 m/s
Using the II equation of motion
S = ut + ½ gt²
Let tₓ be the time where the two stones meet and x be the distance covered from the top of the building
The equation for the stone dropped from top of the building becomes
x = 0 + ½ gtₓ²
The equation for the stone thrown from the base becomes
S - x = U tₓ - ½ gtₓ² (∵ the motion of the stone is in opposite direction)
Adding these two equations,
x + (S - x) = U tₓ
S = U tₓ
200 = 50 tₓ
∴ tₓ = 4 s
Hence, the time after which the two stones meet is tₓ = 4 s
I am pretty sure the answer is C.
Here when an object is placed on the level floor then in that case there are two forces on the object
1). Weight of object downwards (mg)
2). Normal force due to floor which will counterbalance the weight (N)
so when no force is applied on the box at that time normal force is counter balanced by weight.
Now here it is given that A person tried to lift the box upwards
So now there are two forces on the box
1). Applied force of person
2). Normal force due to ground
So now these two forces will counter balance the weight of the crate
So we can write an equation for force balance like

given that

here
m = 30 kg and
g = acceleration due to gravity = 10 m/s^2

now from above equation


So force applied by the person must be 150 N