What you know:
Vi=0m/s
Vf=143.8m/s
A=-9.8m/s
d=???
Use the equation Vf^2=Vi^2+2A(d)
Rearrange to isolate d: d=Vf^2/2A
d=(143.8)^2/2(-9.8)
d=20678.4/-19.6
d=-1055m
The tank was released from a height of 1055m
speed of the car = 27 m/s
speed of truck ahead = 10 m/s
relative speed of car with respect to truck

relative deceleration of car

now the distance before they stop with respect to each other is given by



so it will come at the same speed of truck after 20.6 m distance and hence it will not hit the truck as the distance of the truck is 25 m from car
Part b)
Distance traveled by car before it stops is given by



so it will stop after it will cover total 52.1 m distance
Part c)
time taken by the car to stop



now the distance covered by truck in same time

now after the car will stop its distance from the truck is

<em>so the distance between them is 11.5 m</em>
I’m pretty sure it’s a starting point... unless it’s asking for an actual object
Answer:
1716.75 J
Explanation:
<u>Step </u><u>1</u><u>:</u> First check what we are provided with. As per given question we have:
mass (m) = 70 kg, height (h) = 2.5 m and acceleration due to gravity (g) = 9.81 m/s².
<u>Step</u><u> </u><u>2</u><u>:</u> Check what we are asked to find out.
Work done = Change in Potential energy
The stuff required to solve this question is potential energy. Using the formula: P = mgh. Where P is Potential energy, m is mass, g is acceleration due to gravity and h is height.
<u>Step</u><u> </u><u>3</u><u>:</u> Substitute the known values in the above formula.
→ P = 70 × 2.5 × 9.81
→ P = 1716.75 J
Hence, the work done against the force of gravity is 1716.75 J.