Answer:
1.97 seconds
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.8 m/s²

Solving the above equation we get

So, the time the package was in the air is 1.97 seconds
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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Answer:
The object is dropped, we know the initial velocity is zero. Once the object has left contact with whatever held or threw it, the object is in free-fall. Under these circumstances, the motion is one-dimensional and has constant acceleration of magnitude g.
No so sure
Explanation:
Hope it helps
Its B , the definition of cardiovascular endurance is <span>the ability of the heart, lungs and blood vessels to deliver oxygen to your body tissues.</span>
Answer:
(d) not enough info
Explanation:
because it doesn't specify where the strings are attached
if it was the two ends of the rod then T1 would be equal to T2