Modeling Projectile Motion

First Problem:

Find the maximum height , flight time, and range of a projectile fired from the origin over horizontal ground at an initial speed of 500 m/sec and a launch angle of 60^o. Then graph the path of the projectile.

Answer:

x = 500 cos 60t = 250t
y = 500 sin 60t - 4.9 t² = 250\/3 t - 4.9t²

The maximum height occurs when velocity is 0.
x_v = 250
y_v = 250\/3 - 9.8t
The velocity will be 0 when y_v is 0.
9.8t = 250\/3
t = 250\/3 / 9.8
The height at this time is
y = (250\/3)² / 9.8 - 4.9(250\/3 / 9.8)² = 9566.327 meters

The flight time can be found by finding when y = 0.
250\/3 t - 4.9t² = 0
t(250\/3 - 4.9t) = 0
4.9t = 250\/3
t = 250\/3 / 4.9 = 88.370 sec.

The range is the distance from start to finish horizontally.
x = 250t
At t = 88.370, x = 22,092.485 meters

Second Problem:

To open the 1992 Summer Olympics in Barcelona, bronze medalist archer Antonio Rebollo lit the Olympic torch with a flaming arrow. Suppose that Rebollo shot the arrow at a height of 6 ft above ground level 30 yd from the 70-ft-high cauldron, and he wanted the arrow to reach maximum height exactly 4 ft above the center of the cauldron. Set up the parametric and vector equations.

Answer:

Third Problem:

A baseball is hit when it is 3 ft above the ground. It leaves the bat with initial speed of 152 ft/sec, making an angle of 20^o with the horizontal. At the instant the ball is hit, an instantaneous gust of wind blows in the horizontal direction directly opposite the direction the ball is taking toward the outfield, adding a component of -8.8 ft/sec to the ball’s initial velocity. (8.8 ft/sec = 6 mph)

Answer: