Answer:
the maximum vertical height the person in the cart can reach is 18.42 m
Explanation:
Given;
mass of the person in cart, m₁ = 45 kg
mass of the cart, m₂ = 43 kg
acceleration due to gravity, g = 9.8 m/s²
final speed of the cart before it goes up the hill, v = 19 m/s
Apply the principle of conservation of energy;

Therefore, the maximum vertical height the person in the cart can reach is 18.42 m
A. Using the third equation of motion:
v2 = u2 + 2as
from the question;
the jet was initially at rest
hence u = 0
a = 1.75m/s2
s = 1500m
v2 = 02 + 2(1.75)(1500)
v2 = 5250
v = √5250
v = 72.46m/s
hence it moves with a velocity of 72.46m/s.
b. s = ut + 1/2at2
1500 = 0(t) + 1/2(1.75)t2
1500 × 2 = 2× 1/2(1.75)t2
3000 = 1.75t2
1714.29 = t2
41.4 = t
hence the time taken for the plane to down the runway is 41.4s.
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