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Trajectory Calculator
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Any object thrown in air reaches the earth taking the curved path. This motion is projectile motion and path taken by the object is the trajectory. The range is the maximum distance traveled by the object from initial point. Maximum height is the height attained by the object in air and time of flight is the time taken by the object to reach the final point.
Projectile Motion Formulas
Using the above fig the trajectory equations are given by
Time of flight t = $\frac{2 v_o sin \theta}{g}$
Maximum height reached H =$\frac{v_o^2 sin^2 \theta}{2g}$
Horizontal range, R = $\frac{v_o^2 sin 2\theta}{g}$


Trajectory Calculator is a online tool to calculate the time of flight t, maximum height reached H and horizontal range R. You just have to enter the angle $\theta$ and velocity vo and get the answer instantly.
 

Steps for Trajectory Calculator

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Step 1 :  

Read the problem and check whether angle $\theta$ and release velocity (vo) are given and note it down



Step 2 :  

To find the trajectory parameters use the formulas

Time of flight t = $\frac{2 v_o sin \theta}{g}$



Maximum height reached H = $\frac{v_o^2 sin^2 \theta}{2g}$



Horizontal range, R = $\frac{v_o^2 sin 2\theta}{g}$




Substitute the values in above formula and get the answer.



Problems on Trajectory Calculator

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  1. A bullet is fired with the speed of 85 m/s at an angle of 24o and hits the target. Calculate the range.


    Step 1 :  

    Given: Initial velocity (vo) = 85 m/s, angle ($\theta$) = 24o. To find the range R



    Step 2 :  

    The range of a projectile motion is given by
    R = $\frac{v_o^2 sin 2 \theta}{g}$
       = $\frac{85^2 sin 48^o}{9.8}$
       = $\frac{85^2 \times 5369.22}{9.8}$
       = 547.88 m



    Answer  :  

    Range of projectile R = 547.88 m.



  2. A ball is projected in air has the velocity of 44 m/s at the angle of 30o. What will be its time of flight and maximum height reached?


    Step 1 :  

    Given: Initial velocity (vo) = 44 m/s, angle $\theta$ = 30o. To find maximum height H and time of flight t



    Step 2 :  

    Hence the maximum height reached is
    H = $\frac{v_o^2 sin^2 \theta}{2g}$



       = $\frac{44^2 sin^2 30^o}{2 \times 9.8}$



       = $\frac{484}{19.6}$



       = 24.69 m

    Time of flight is
    t = $\frac{2V_o sin \theta}{g}$



      = $\frac{2 \times 44 \times sin 30^o}{9.8}$



      = 4.489 s.



    Answer  :  

    Maximum height H = 0.56 m and time of flight t = 4.489 s.



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