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, while the horizontal component stays unchanged because there is no horizontal acceleration.
The angle with the vertical is 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)