Derivation of Position-Velocity Relation by Graphical Method
A stone is th...
Question
A stone is thrown vertically upwards. When the stone is at a height equal to half of its maximum height its speed will be 10 m/s, then the maximum height attained by the stone is (Take g =10 m/s2)
A
3 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
15 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
1 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
10 m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D 10 m Let u be the initial velocity and h be the maximum height attained by the stone. v21=u2−2gh, (10)2=u2−2×10×h2 100=u2−10h ....(i) Again at height h, v22=u2−2gh (0)2=u2−2×10×h u2=20h ... (ii) So, from Eqs. (i) and (ii) we have 100=10h h=10m