SOLUTION: The range of a projectile is the horizontal distance it travels before it reaches the ground. The greatest range is achieved if the projectile is thrown at 45° to the horizontal.

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Question 1200490: The range of a projectile is the horizontal distance it travels before it reaches the ground. The greatest range is achieved if the projectile is thrown at 45° to the horizontal.
A ball is thrown with an initial velocity of 40 ms^-1, Calculate its greatest possible range when air resistance is considered to be negligible.
Answer by ikleyn(52776) About Me  (Show Source):
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The range of a projectile is the horizontal distance it travels before it reaches the ground.
The greatest range is achieved if the projectile is thrown at 45° to the horizontal.
A ball is thrown with an initial velocity of 40 ms^-1, Calculate its greatest possible range
when air resistance is considered to be negligible.
~~~~~~~~~~~~~~~~~ 

So, the angle to the horizon of the initial velocity is assumed to be 45°.


At this condition, the horizontal component of the velocity is 

    V%5Bhor%5D = 40*cos(45°) = 40*(sqrt(2)/2) = 40*0.707 = 28.285 m/s;


The vertical component of the velocity is

    V%5Bvert%5D = 40*sin(45°) = 40*(sqrt(2)/2) = 40*0.707 = 28.285 m/s  

(has the same numerical value as the horizontal component).


The flight time is  t = 2%2A%28V%5Bvert%5D%2Fg%29 = 2%2A%2828.285%2F9.81%29 = 5.767 seconds (approximately).


    Here g = 9.81 m/s^2  is the gravity acceleration at the Earth surface.


The horizontal distance to fly (the range) is  V%5Bhor%5D%2At = 28.285*5.767 = 163.120 m (rounded).


ANSWER.  The greatest possible range is approximately 163.120 m under given conditions.

Solved.