A boy playing on the roof of a 10m high building throws a ball with a speed of 10 m/s at…
Physics · JEE Main · NTA Exams — Kinematics
A boy playing on the roof of a 10m high building throws a ball with a speed of 10 m/s at an angle of 30o with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground?
(g = 10 m/s2, sin 30o = \(\frac{1}{2},\)\(\left.\cos 30^{\circ}=\frac{\sqrt{3}}{2}\right)\)
(g = 10 m/s2, sin 30o = \(\frac{1}{2},\)\(\left.\cos 30^{\circ}=\frac{\sqrt{3}}{2}\right)\)
- 8.66 m
- 5.20 m
- 4.33 m
- 2.60 m
Answer
(A) 8.66 m
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