Projectile Motion Using Coordinates

  • Thread starter i8mysocks
  • Start date
  • #1
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1. A plane with velocity (350 m/s, 40 deg) dropped a rescue package at location (0, 300 m) and the package landed at location (range, 50m). Find the range and final velocity.



2. d = vsin(40) x t + (1/2)x(a)x(t^2)
dx = vcos(40) x t




3. final velocity:
first find the time in air:
300 m= 350sin(40) x t + 1/2 (10 m/s^2)(t^2)
t = 15 s
final velocity = 350sin(40) + (10 m/s^2)(15s)

300 m = t ((350sin(40) + (5 m/s^2)(t)) let t = 0
t = 15 s

final velocity = 350sin(40) + (10 m/s^2)(15s)
= 375 m/s

horizontal distance = 350cos(40) x (15 s)
= 4021.73 m

using coordinates (0, 300) and (?, 50) solve for ? using the distance formula:

4021.73 m = sq root( (300-50)^2 + (0 - ?)^2))
(4021.73 m)^2 = (300-50)^2 +(0-?)^2
(1.6 x 10^7m) - (300 - 50)^2 = (-?)^2
sq root (1.59375 x 10^7) = ?
? = 3392.18

...basically I would like to know if my math was correct and if I used the correct coordinates to solve for "?". Thank you for your help! :D
















 
Last edited:

Answers and Replies

  • #3
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You really think so? The coordinates I used to solve for "range" were correct, right? By the way, thank you very much for taking the time to look at this problem. I really appreciate it :D
 

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