Example 5: Horizontal Distance Traveled by a Projectile
In this video, I determine the vector formula for path of a projectile launched at an angle by integrating the acceleration due to gravity acting on it, and then integrating its velocity vector. When the y-component of the position vector is equal to zero (i.e. the object strikes the ground), the corresponding non-trivial time variable corresponds to the horizontal distance traveled. This distance is maximized when the launch angle is π/4 radians or 45 degrees.
#math #vectors #calculus #physics #education
Timestamps
- Example 5: Find angle that gives the maximum horizontal distance traveled by a projectile – 0:00
- Solution: Gravity is the only force that acts on the object – 0:49
- Integrate the acceleration due to gravity to get the velocity vector – 3:02
- Integrate the velocity vector to obtain the position vector – 5:29
- Components of the initial velocity – 8:04
- Position vector and the parametric equations of trajectory – 11:00
- The projectile follows the path of a parabola since y is a quadratic function of x – 13:39
- Horizontal distance is the value of x when y = 0 – 16:23
- Solving for the distance by plugging in the non-trivial t-value associated with y = 0 – 18:57
- Recall the sine double-angle formula – 21:17
- The distance is maximum when the launch angle is 45 degrees of π/4 radians – 21:42 .
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