The question lacks details. Here is the full question.
The accompanying image was captured with a camera capable of shooting between one and two frames per second. A series of photos was merged into this single image, meaning the vehicles depicted are actually the same car, documented at different intervals.
Assuming the camera produced 1.3 frames per second for this image and that the length of the car is approximately 5.3 meters, based on this information and the photo, how fast was the car moving?
Answer: v = 6.5 m/s
Explanation: The problem requires calculating the car's velocity. Velocity can be computed using:

Since the camera captured 7 images of the car and its length is noted as 5.3, the car's displacement is:
Δx = 7(5.3)
Δx = 37.1 m
The camera operates at 1.3 frames per second and recorded 7 images, thus the time driven by the car is:
1.3 frames = 1 s
7 frames = Δt
Δt = 5.4 s
<pconsequently the="" car="" was="" driving="" at:="">

v = 6.87 m/s
<pthe car="" moved="" at="" an="" estimated="">velocity of
6.87 m/s.
</pthe></pconsequently>
Answer:
h = 10 m
Explanation:
Given,
mass of the platform = 50 Kg
Kinetic energy = 5000 J
height from which the diver dove =?
Taking the acceleration due to gravity as 10 m/s²
Using the conservation of energy principle
Kinetic energy is transformed into potential energy
K.E = P.E
K.E = m g h
5000 = 50 x 10 x h
500 h = 5000

h = 10 m
The height from which the diver dove is equal to h = 10 m