Incomplete question as time is missing.I have assumed some times here.The complete question is here
Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, (d) 2.00 s, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 10 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.
Explanation:
Given data
Vi=10 m/s
S=70 m
(a) t₁=0.5 s
(b) t₂=1 s
(c) t₃=1.5 s
(d) t₄=2 s
(e) t₅=2.5 s
To find
Displacement S from t₁ to t₅
Velocity V from t₁ to t₅
Solution
According to kinematic equation of motion and given information conclude that v is given by

Also get the equation of displacement

These two formula are used to find velocity as well as displacement for time t₁ to t₅
For t₁=0.5 s

For t₂
For t₃

For t₄

For t₅

Answer:
: Rocket weight on earth
: Rocket weight on moon
Explanation:
Conceptual analysis
Weight is the force with which a body is attracted due to the action of gravity and is calculated using the following formula:
W = m × g Formula (1)
W: weight
m: mass
g: acceleration due to gravity
The mass of a body on the moon is equal to the mass of a body on the earth
The acceleration due to gravity on a body is different on the moon and on the earth
Equivalences
1 slug = 14.59 kg
Known data



Problem development
To calculate the weight of the rocket on the moon and on earth we replace the data in formula (1):
: Rocket weight on earth
: Rocket weight on moon
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The short answer is, maybe.
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Answer
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