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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

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

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

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