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Derive the formula of range of projectile.Please explain in simple way. Thanks 🙏🙏🙏

Derive the formula of range of projectile.Please explain in simple way. Thanks 🙏🙏🙏

Grade:9

1 Answers

Arun
25750 Points
4 years ago
 

Velocity along x-axis = ux = u cos θ

Acceleration along x-axis ax = 0

Velocity along y-axis = uy = u sin θ

Acceleration along y-axis ay = -g

Here we use different equation of motions of one dimension derived earlier to get the different parameters.

\vec{v}=\vec{v_{0}}-\vec{g}t                           …... (a) 

\vec{y}-\vec{y_{0}}=\vec{v_{0}}t-\frac{1}{2}\vec{g}t^{2}           …... (b)         

v2 = v02 – 2g (y-y0)                      …... (c) 
 

Total Time of Flight

When body returns to the same horizontal level, the resultant displacement in vertical y-direction is zero. Use equation b.

Therefore, 0 = (u sin θ) t - (½)gt2,

As t cannot equal to zero, then, total time of flight,

Horizontal Range

Projectile Motion

Horizontal Range (OA=X)   = Horizontal velocity × Time of flight

= u cos θ × 2 u sin θ/g

So horizontal range,

Maximum Height

At the highest point of the trajectory, vertical component of velocity is zero.

Therefore, 0 = (u sin θ)2 - 2g Hmax

So, maximum height would be,

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