An athlete throw a jevelin a maximum distance of 80 m How long is it in air and to what flight does it rises
Answers
Explanation:
Hint: to find the solution to this question consider the equations of projectile motion. Find the equation of projectile motion. Put the values on the equation range, find the angle and then put these values on the equation of maximum height to find our answer.
Formula used:
The horizontal range for a projectile motion is given by, R=u2sin2θg
The maximum horizontal range is given the mathematical expression as,
hmax=u2sin2θ2g
Complete step by step answer:
Projectile motions can be defined as the motion of an object near the earth's surface, which will move under the gravity to finally fall on earth. This trajectory has both vertical and horizontal components.
The horizontal range for a projectile motion is given by, R=u2sin2θg
Where, u is the initial velocity given to the object, g is the acceleration due to gravity and θ is the angle with the horizontal at which the object is thrown.
Now, range will be maximum when the value of sinθ will be maximum.
So,
2θ=900θ=9002=450
So, maximum range is,
R=u2g=80m
Now, the maximum horizontal range is given the mathematical expression as,
hmax=u2sin2θ2g
Putting the values in the above equation, we can find that,
hmax=12u2gsin2θhmax=12×80m×sin2450hmax=12×80m×(12–√)2hmax=12×80m×12hmax=20m
So, the maximum height to which it will rise is 20 m.
The correct option is (B).
Note: The maximum horizontal range and the maximum height of an object in projectile motion is always dependent on the angle the object is thrown. With the same initial velocity also, the maximum ranges will change if the direction of the velocity or the angle with the horizontal is changed.