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A ball is projected with a velocity, 10ms−1, at an angle of 60∘ with the vertical direction. Its speed at the highest point of its trajectory will be
- a)Zero
- b)5√3ms−1
- c)5ms−1
- d)10ms−1
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A ball is projected with a velocity, 10ms−1, at an angle of 60∘ with ...
Given, initial velocity of the ball, u = 10 m/s
Angle of projection, θ = 60°
To find: speed of the ball at the highest point of its trajectory
At the highest point of the trajectory, the vertical component of velocity becomes zero. The horizontal component of velocity remains constant throughout the motion.
Horizontal and vertical components of velocity can be found as:
uₓ = u cosθ = 10 cos60° = 5 m/s
uᵧ = u sinθ = 10 sin60° = 5√3 m/s
At the highest point, vᵧ = 0
Using the equation of motion vᵧ = uᵧ - gt, where g is acceleration due to gravity (9.8 m/s²):
0 = 5√3 - gt
t = 5√3/9.8 seconds
Using the equation of motion vₓ = uₓ, since there is no acceleration in the horizontal direction:
vₓ = uₓ = 5 m/s
The speed of the ball at the highest point of its trajectory can be found using Pythagoras theorem:
v = √(vₓ² + vᵧ²)
v = √(5² + (5√3)²)
v = √(25 + 75)
v = √100
v = 10 m/s
Therefore, the correct answer is option B) 5√3 m/s.
Angle of projection, θ = 60°
To find: speed of the ball at the highest point of its trajectory
At the highest point of the trajectory, the vertical component of velocity becomes zero. The horizontal component of velocity remains constant throughout the motion.
Horizontal and vertical components of velocity can be found as:
uₓ = u cosθ = 10 cos60° = 5 m/s
uᵧ = u sinθ = 10 sin60° = 5√3 m/s
At the highest point, vᵧ = 0
Using the equation of motion vᵧ = uᵧ - gt, where g is acceleration due to gravity (9.8 m/s²):
0 = 5√3 - gt
t = 5√3/9.8 seconds
Using the equation of motion vₓ = uₓ, since there is no acceleration in the horizontal direction:
vₓ = uₓ = 5 m/s
The speed of the ball at the highest point of its trajectory can be found using Pythagoras theorem:
v = √(vₓ² + vᵧ²)
v = √(5² + (5√3)²)
v = √(25 + 75)
v = √100
v = 10 m/s
Therefore, the correct answer is option B) 5√3 m/s.
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Community Answer
A ball is projected with a velocity, 10ms−1, at an angle of 60∘ with ...
At highest point vertical component of velocity become zero.
At highest point speed of object = 10 cos30∘
= 5√3m/s
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