Answer:
a = 1.152s
b = 0.817 m
c = 7.29m/s
Explanation: let the following
From the first equation of linear motion
V = u+at..........1
parameters be represented as :
t = Time taken
v = Final velocity
a = Acceleration due to gravity = 9.8m/s²
u = Initial velocity = 4 m/s
s = Displacement
V = 0
Substitute the values into equation 1
0 = 4-9.8(t)
-4 = -9.8t
t = 4/9.8
t = 0.408s
From : s = ut+1/2at^2.........2
S = 4×0.408+0.5(-9.8)×0.408^2
S= 1.632-4.9(0.166)
S = 1.632-0.815
S = 0.817m
Her highest height above the board is 0.817 m
Total height she would fall is 0.817+1.90 = 2.717 m
From equation 2
s = ut+1/2at^2
2.717 m = 0t+0.5(9.8)t^2
2.717 m = 0+4.9t^2
2.717 m = 4.9t^2
2.717/4.9 = t^2
0.554 =t^2
t =√0.554
t = 0.744s
Hence, her feet were in the air for 0.744+0.408seconds
= 1.152s
Also recall from equation 1
V= u+at
V = 0+9.8(0.744)
V = 7.29m/s
Hence, the velocity when she hits the water is 7.29m/s
Finally,
a = 1.152s
b = 0.817 m
c = 7.29m/s
Explanation:
It is given that,
A particle starts from rest and has an acceleration function as :
(a) Since, 
v = velocity
(b) 
x = position
(c) Velocity function is given by :
t = 1 seconds
So, at t = 1 second the velocity of the particle is zero.
Answer:
350J
Explanation:
Given parameters:
Weight of bag = 20N
Distance moved horizontally = 35m
Force applied = 10N
Unknown:
Work done on the bag = ?
Solution:
Work done is the force applied to move a body through given distance.
Work done = Force applied x distance
So;
Work done = 10 x 35 = 350J
The rock's height <em>y</em> at time <em>t</em> is given by
<em>y</em> = 45 m + (7.2 m/s) <em>t</em> - 1/2 <em>g</em> <em>t </em>²
where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Set <em>y</em> = 0 and solve for <em>t</em> :
0 = 45 m + (7.2 m/s) <em>t</em> - 1/2 <em>g</em> <em>t </em>² → <em>t</em> ≈ 3.9 s
