An ice skier is sliding from a point A as shown in the figure below. If the slider starts from rest and acceleration due to gravity is 10m/s2, then what is the velocity of the skier at point B.
A
10√5m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
10√2m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
15√5m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
20√2m/s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D20√2m/s On applying the principle of energy conservation, loss in potential energy of skier equals gain in the kinetic energy of the skier. hA= Height of A from the ground =50m hB= Height of B from the ground=10m
Thus, (mghA−mghB)=(12mV2B−12mV2A)
As the initial velocity at point A is zero mg(hA−hB)=12mV2B ⟹g(50−10)=12V2B ⟹12V2B=(10)×(40) ⟹V2B=2×400=800 ⟹VB=√800=20√2m/s
Hence, the speed of the skier at point B is 20√2m/s