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
The answer is B 1,500 meters since 1 kilometer to meters is 1,000 and u add the 500
Answer:
Total displacement will be 47 meter
Total distance will be 83 meters
Explanation:
We have given that first the student go eastward towards bus stop 20 meters
But he realizes that she dropped his physics notebook and so h=she turns back along the same way up to 18 meters
So displacement = 20-18 = 2 meters
And he travel 45 meters in east along the bus stop so total displacement = 45+2 = 47 meters
Total distance traveled by the student = 20+18+45 = 83 meters
Answer:
sorry about the other person but its b
Explanation:
