An athlete throws a javelin with velocity of 20mper sec at an angle of 15 degree with the horizontal. How far the javelin would hit the ground from the point of projection.
Answers
Answer:
Explanation:range of javeline,
R=
(u^2)(sin2Φ)/g
i.e.{(20) ^2 }sin 2 * 15 ° / 9.8
=400sin30°/9.8
=(400 * 1/2 )/ 9.8
=20.4 m
Therefore the javelin would hit the ground after 20.387 m from the point of projection.
Given:
Velocity of the javelin = 20 m/s²
Angle made by the javelin with the horizontal = 15°
To Find:
The horizontal distance or range of the javelin.
Solution:
The given question can be solved as shown below.
Given that,
Velocity of the javelin ( u ) = 20 m/s²
Angle made by the javelin with the horizontal ( θ ) = 15°
The range or the horizontal distance traveled by an object in the projectile motion is given by,
⇒ R = ( u² × Sin 2θ ) / g
where g = acceleration due to gravity = 9.81 m/s²
⇒ R = ( 20² × Sin ( 2. 15 ) ) / 9.81
⇒ R = ( 400 × Sin 30 ) / 9.81
⇒ R = 200 / 9.81
⇒ R = 20.387
Therefore the javelin would hit the ground after 20.387 m from the point of projection.
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