Answer:
After 1 sec = 4.9 m
After 2 sec = 19.6 m
After 3 sec = 44.1 m
After 4 sec = 78.4 m
After 5 sec = 122.5 m
Explanation:
After 1 sec:
<em>u=0m/s t=1 s a=9.8m/s²</em>
s = ut + (1/2)at²
=0(1) + (1/2)(9.8)(1²) = 4.9m
After 2 sec:
<em>u=0m/s t=2 s a=9.8m/s²</em>
s = ut + (1/2)at²
=0(2) + (1/2)(9.8)(2²) = 19.6m
After 3 sec:
<em>u=0m/s t=3 s a=9.8m/s²</em>
s = ut + (1/2)at²
=0(3) + (1/2)(9.8)(3²) = 44.1m
After 4 sec:
<em>u=0m/s t=4 s a=9.8m/s²</em>
s = ut + (1/2)at²
=0(4) + (1/2)(9.8)(4²) = 78.4m
After 5 sec:
<em>u=0m/s t=5 s a=9.8m/s²</em>
s = ut + (1/2)at²
=0(5) + (1/2)(9.8)(5²) = 122.5m
Answer:
Angle is 55.52°
and Initial Speed is v=26.48 m/s
Explanation:
Given data
Applying the kinematics equations for motion with uniform acceleration in x and y direction
So
Put the value of v₀ from equation (1) to equation (2)
So
Put that angle in equation (1) or equation (2) to find the initial velocity
So from equation (1)
Answer:
Explanation:
The time period of geosynchronous satellite must be equal to T .
The radius of its orbit will be ( R+ h )
orbital velocity V₀ = 
Time period T = 2π( R + h ) / V₀
= 2π( R + h ) x 
= R +h
h = - R.
Answer:
<u>We are given:</u>
displacement (s) = 130 m
acceleration (a) = -5 m/s²
final velocity (v) = 0 m/s [the cars 'stops' in 130 m]
initial velocity (u) = u m/s
<u>Solving for initial velocity:</u>
From the third equation of motion:
v² - u² = 2as
replacing the variables
(0)² - (u)² = 2(-5)(130)
-u² = -1300
u² = 1300
u = √1300
u = 36 m/s
