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MATHEMATICS - CASE STUDY SOLUTION


Solution Diagram

Table: Data for the Trajectory
t (s)
x (ft)
y (ft)
0
0
0
5
1,000
1,330
10
2,000
1,854
15
3,000
1,574
20
4,000
488
21.516
4,303
0


Trajectory of the Projectile

 

Recall, the path of the projectile was given as by the following parametric equations:

     x = (Ucosθ)t
     y = (Usinθ)t - 0.5gt2

For U = 400 ft/s and θ = 60o, the parametric equations reduce to:

     x = (400 cos 60)t = 200t
     y = (400 sin 60)t - 0.5(32.2)t2 = 346.4t - 16.1t2

When the projectile hits the ground, y is zero. Thus,

     346.4t - 16.1t2 = 0
     16.1t2 = 346.4t
     t = 346.4/16.1 = 21.52 s

The corresponding x-position of the projectile is

     x = 200(21.52) = 4,304 ft

Based on the parametric equations, the positions of the trajectory are determined by varying the time t. The trajectory of the projectile is plotted in the figure shown on the left.