Apply this kinematics equation for the particle's vertical motion:
D = Vt + 0.5At²
D = vertical distance traveled, t = time, V = initial vertical velocity, A = vertical acceleration
Given values:
D = 49m
V = 34.3m/s
A = -9.81m/s²
Plug in and solve for t:
49 = 34.3t - 4.905t²
4.905t² - 34.3t + 49 = 0
Use the quadratic roots formula to find the values of t:
t ≈ 2, 5
The particle reaches a height of 49m at t = 2s & t = 5s
Answer: 60 m/s
Explanation:
KE = 1/2mv^2
v = √ (2KE / m)
v = √ (2 x 36 J)/(0.02 kg)
v = √ 72 J/0.02 kg
v = √ 3600 m/s
v = 60 m/s
Answer:
6844.5 m/s.
Explanation:
To get the speed of the satellite, the centripetal force on it must be enough to change its direction. This therefore means that the centripetal force must be equal to the gravitational force.
Formula for centripetal force is;
F_c = mv²/r
Formula for gravitational force is:
F_g = GmM/r²
Thus;
mv²/r = GmM/r²
m is the mass of the satellite and M is mass of the earth.
Making v the subject, we have;
v = √(GM/r)
We are given;
G = 6.67 × 10^(-11) m/kg²
M = 5.97 × 10^(24) kg
r = 8500 km = 8500000
Thus;
v = √((6.67 × 10^(-11) × (5.97 × 10^(24)) /8500000) = 6844.5 m/s.
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