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A ball is projected with a velocity 10 m/s at an angle 30 degree with the horizontal direction. The speed of ball at highest point of its trajectory will be (1) Zero (2) 5√3 m/s (3) 5 m/s (4) 10 m/s Answer is option (2) 5√3 m/s?
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A ball is projected with a velocity 10 m/s at an angle 30 degree with ...
Explanation:
When a ball is projected at an angle with horizontal, it follows a parabolic path. At the highest point of its trajectory, the vertical component of its velocity becomes zero while the horizontal component remains constant.
Step 1: Find the initial velocity of the ball in horizontal and vertical direction.
The given velocity is 10 m/s at an angle of 30 degrees.
Horizontal component of velocity, Vx = V cos θ = 10 cos 30° = 5√3 m/s
Vertical component of velocity, Vy = V sin θ = 10 sin 30° = 5 m/s
Step 2: Find the time taken by the ball to reach the highest point.
At highest point, the vertical component of velocity becomes zero.
Using the equation, v = u + at, where v is final velocity, u is initial velocity, a is acceleration and t is time,
0 = 5 - 9.8t
t = 5/9.8 seconds
Step 3: Find the velocity of the ball at the highest point.
At the highest point, the velocity of the ball is equal to the horizontal component of velocity.
Using the equation, v = u + at, where v is final velocity, u is initial velocity, a is acceleration and t is time,
v = 5√3 + (0)(5/9.8)
v = 5√3 m/s
Therefore, the speed of the ball at the highest point of its trajectory is 5√3 m/s.
Answer: Option (2) 5√3 m/s.
When a ball is projected at an angle with horizontal, it follows a parabolic path. At the highest point of its trajectory, the vertical component of its velocity becomes zero while the horizontal component remains constant.
Step 1: Find the initial velocity of the ball in horizontal and vertical direction.
The given velocity is 10 m/s at an angle of 30 degrees.
Horizontal component of velocity, Vx = V cos θ = 10 cos 30° = 5√3 m/s
Vertical component of velocity, Vy = V sin θ = 10 sin 30° = 5 m/s
Step 2: Find the time taken by the ball to reach the highest point.
At highest point, the vertical component of velocity becomes zero.
Using the equation, v = u + at, where v is final velocity, u is initial velocity, a is acceleration and t is time,
0 = 5 - 9.8t
t = 5/9.8 seconds
Step 3: Find the velocity of the ball at the highest point.
At the highest point, the velocity of the ball is equal to the horizontal component of velocity.
Using the equation, v = u + at, where v is final velocity, u is initial velocity, a is acceleration and t is time,
v = 5√3 + (0)(5/9.8)
v = 5√3 m/s
Therefore, the speed of the ball at the highest point of its trajectory is 5√3 m/s.
Answer: Option (2) 5√3 m/s.
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A ball is projected with a velocity 10 m/s at an angle 30 degree with the horizontal direction. The speed of ball at highest point of its trajectory will be (1) Zero (2) 5√3 m/s (3) 5 m/s (4) 10 m/s Answer is option (2) 5√3 m/s?
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