A ball is projected with a velocity, 10 m s-1, at an angle of 60 with the vertical direction. Its speed at the highest point of its trajectory will be
- Zero
- 5 3 m s-1
- 5 m s-1
- 10 m s-1
Answer
(2) 5 3 m s-1
Solution
At the highest point the vertical component of velocity becomes zero; the horizontal component stays unchanged (no horizontal acceleration).
Angle with the vertical = 60 , so the horizontal component is:
vx = u 60
vx = 10 32
vx = 5 3 m s-1
Speed at the highest point = 5 3 m s-1
Answer: (2)