Answer:
h> 2R
Explanation:
For this exercise let's use the conservation of energy relations
starting point. Before releasing the ball
Em₀ = U = m g h
Final point. In the highest part of the loop
Em_f = K + U = ½ m v² + ½ I w² + m g (2R)
where R is the radius of the curl, we are considering the ball as a point body.
I = m R²
v = w R
we substitute
Em_f = ½ m v² + ½ m R² (v/R) ² + 2 m g R
em_f = m v² + 2 m g R
Energy is conserved
Emo = Em_f
mgh = m v² + 2m g R
h = v² / g + 2R
The lowest velocity that the ball can have at the top of the loop is v> 0
h> 2R
Speed =distance/time
3.25=3.00/time
3.25xt=3.00
t=3/3.25
s=0.9s
Answer:
Total time taken=110 seconds
Total distance traveled=480m
Explanation:
First of all, we find the total time taken:
For that, we use the formula : Distance/Speed= Time
Time for part 1 : 200/5=40 seconds
Time for part 2 : 280/4=70seconds
Total time taken=110 seconds
Total distance traveled=480m
Average Speed= 480/110=4.36 m/s
Total displacement=200-280=-80m (Since this is displacement, we need to find the distance between the initial and final point. Also, I've taken east direction as positive and west as negative)
Average Velocity=-80/110=-0.72 m/s
OR 0.72m/s towards west.
