Using V²=u² +2gH
let's assume the ball falls from rest, then it's initial velocity = 0ms-¹ (U=0)
then from the above info V= 35ms-¹
g= 10ms-² (value of acceleration due to gravity to the nearest tens)
so input your values above into the given formula to find height
35² = 0² +2×10×H
1225 = 20H
H = 1225÷20 = 61.25m
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Gravitational Potential Energy = weight x height
for 1 meter:
GPE = 5 x 1
= <u>5N</u>
for 1.5 metres:
GPE = 5 x 1.5
= <u>7.5N</u>
for 2 metres:
GPE = 5 x 2
= <u>10N</u>
Answer:
x(t)=0.337sin((5.929t)
Explanation:
A frictionless spring with a 3-kg mass can be held stretched 1.6 meters beyond its natural length by a force of 90 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 2 m/sec, find the position of the mass after t seconds.
Solution. Let x(t) denote the position of the mass at time t. Then x satisfies the differential equation
Definition of parameters
m=mass 3kg
k=force constant
e=extension ,m
ω =angular frequency
k=90/1.6=56.25N/m
ω^2=k/m= 56.25/1.6
ω^2=35.15625
ω=5.929
General solution will be
differentiating x(t)
dx(t)=-5.929c1sin(5.929t)+5.929c2cos(5.929t)
when x(0)=0, gives c1=0
dx(t0)=2m/s gives c2=0.337
Therefore, the position of the mass after t seconds is
x(t)=0.337sin((5.929t)