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. |