Answer:
Explanation:
Horizontal displacement
x = 120 t
Vertical position
y = 3610 - 4.9 t²
y = 0 for the ground
0 = 3610 - 4.9 t²
t = 27.14 s
This is the time it will take to reach the ground .
During this period , horizontal displacement
x = 120 x 27.14 m
= 3256.8 m
So packet should be released 3256.8 m before the target.
Answer:
Explanation:
Givens
vi = 0
a = 9.81
d = 4.50 m
vf = ?
Formula
vf^2 = vi^2 + 2 * a * d
Solution
Substitute the knowns into the formula
vf^2 =0 + 2 * 9.81 * 4.50
vf^2 = 88.29 Take the square root of both sides.
sqrt(vf^2) = sqrt(88.29)
vf = 9.40 m/s
Answer:
<em>17 m/s west</em>
Explanation:
Runner 1 has velocity = 10 m/s west
runner 2 has velocity = 7 m/s east
From the frame of reference of runner 2, we can imagine runner 2 as standing still, and runner 1 moving away from him, towards the west with their combined velocity of
velocity = 10 m/s + 7 m/s = <em>17 m/s west</em>
