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heyItsLuna234
01.12.2020 •
Mathematics
Calculus application
1. Supposed that a ball is dropped from the upper observation deck of the Horizons 101 Tower 1., located
178 m above the ground. (a) What is the velocity of the ball after 5 sec? (b) How fast is the ball traveling
when it hits the ground?
Solved
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Ответ:
49m/s
59.07 m/s
Step-by-step explanation:
Given that :
Distance (s) = 178 m
Acceleration due to gravity (a) = g(downward) = 9.8m/s²
Velocity (V) after 5 seconds ;
The initial velocity (u) = 0
Using the relation :
v = u + at
Where ; t = Time = 5 seconds ; a = 9.8m/s²
v = 0 + 9.8(5)
v = 0 + 49
V = 49 m/s
Hence, velocity after 5 seconds = 49m/s
b) How fast is the ball traveling when it hits the ground?
V² = u² + 2as
Where s = height = 178m
V² = 0 + 2(9.8)(178)
V² = 0 + 3488.8
V² = 3488.8
V = √3488.8
V = 59.07 m/s
Ответ:
BENG SHE HANAU
Step-by-step explanation: