Answer:
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
Hello:
ln(2e^3x) =ln2+lne^3x
<span> =ln2 +3xlne
= ln2+3x .....lne=1
</span><span>choice B</span>
D because you need to make the denominators have a common multiple.
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB I think
Answer:
40...................................
