If a projectile is projected at an angle 0 from the horizontal with velocity u under the gravitational field of the earth, prove that the path of the projectile will be a parabola !
Projectile Given Angular Projection
![](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s2aSieXicIfjOEvR-37rWbONkS9qqsh295rhw7vY_srXU53tZgGODwTeVyeH0AIhr5kJTCE07YMuOMvKlcL4eLhw_Oih1vkLVs191k_PZEUXriQO6XYfxpNbzvItTO2hcB84VngyDOOQXh92e0wCifZJdLPyp6_IUmQee9hGksxCeN3XkFpVZnN7e2H5S4jehY651AbtgFRIxTFopcB6YKgnfh1zY=s0-d)
Here, x0 = 0, ux = u cos θ, ax = 0
[
Velocity of an object in the horizontal direction is constant]
Putting these values in equation (i),
⇒ x = ut cos θ
⇒
For motion along vertical direction, the acceleration ay is −g.
The position of the object at any time t along the vertical direction is given by,
Here,![](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t2myR4CRxj_ktT9uNP0uQGGfykrSpJLagK9ZStGRQG9_bFjDexQC09aahXuHARXS-ULXuKKQW_-B_pJJQK4L8C1O_jfetVPHyZ86Nd1RoDBAcz_S9tSciSGpSZD75QQQJJuE5BZiHjz-9eYLhaM6gIYUCnR-SJNVgqQ0mV4LUqcbi2HFcqUG-tVP_vqttGNU1pMXGu3-wZ_7ivb8AOE7eNC79Bz_O6=s0-d)
![](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tCIVriJ97MziTO1XZkKlGCV44yL8goiFRTO6ZGHLSzTXARQyH-U0VfjmMn7ijiTr51pAeD6gQ7j2FyYMXcQb1U365LYqkgxuqYJGPXxpzkZjM9L9QHhNNRRfKn29pK6azBwYzW9UdR72v0RMAl80SxS_Lzb7XVNOb8MYQVoJI6VyzUVpB_tSObuJABYBDS0m4Ddeuhiu0fE6YwZVnS2MuM2O1WMkA=s0-d)
∴![](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tv1W3V2z8fOtgDj2kjWwjWigvX-djVT6qw2p1-I2vhau9dOHNLePCWPQaR34UmCFOupr8UrJIhPXJkRreqCTwN5sae39vYClBFicS_BtjaSPZlmXIQU6z251T0ncf_-SS9TbI1x1LSsgTCSC4F7chzapUs9cXJfh0zmQIprdz0xn7msMzvBXcaC780oG6zQThVw6E1bHyZt1W1pY6T-Yt7zAwzp2o=s0-d)
Putting the value of t from equation (ii),
![](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vqoW3wrSJdpCiJp8oUXUm9i8_3p8UPU-MdEUjQrgzWY9tg6PG5e9lwfVD3d1C02GTkoKkZBVyW8vjjaIxOgqyGsyyRn9fk5fS4_dpFOWqIk-d0o4PpXxAwjtQA5mhOngZKzmXeSeZGP0qy-h-LCuyJKY6MIh5OMYgWBjplsd0wyy9_fR4f0mJ6wwnnyXCnxYFPzwHUxPc8PrMq1BCDBLD2m7LBz7Q=s0-d)
⇒![](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vfWv7pF_3XxO606wgjD6YVTlMLiT2gqFV-tiNJGZ0GV0ItKQcV4fYCcRi4Miukvp0szuLY7jXtCqC82AUF_qizMBZ47teJfx5NYpCYLjvImzLefLO3xMt9O9qig5I5u0Na8fmFbDdcqMW5WaBIHz3OqEgWM1uUsRSt15ZMi2vMuYcY-e4MWiOfBYJczcfS5Ey-Jhbrnugf6bC9fWaDpKrMoR0gyoc=s0-d)
This is an equation of a parabola. Hence, the path of the projectile is a parabola.
- Equation of path of projectile − Suppose at any time t, the object is at point P (x, y).
Here, x0 = 0, ux = u cos θ, ax = 0
[
Putting these values in equation (i),
⇒ x = ut cos θ
⇒
For motion along vertical direction, the acceleration ay is −g.
The position of the object at any time t along the vertical direction is given by,
Here,
∴
Putting the value of t from equation (ii),
⇒
This is an equation of a parabola. Hence, the path of the projectile is a parabola.
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