KE = 1/2 * m * v^2
KE = 1/2 * 0.135 * 40^2
KE = 1/2 * 0.135 * 1600
KE = 108 J
M=2.45 because you multiply out the equation on the right and divide by 10
Answer:
Hope it helped
stay safe, mark BRAINLIEST
Answer:
The velocity with which the jumper strike the mat in the landing area is 6.26 m/s.
Explanation:
It is given that,
A high jumper jumps over a bar that is 2 m above the mat, h = 2 m
We need to find the velocity with which the jumper strike the mat in the landing area. It is a case of conservation of energy. let v is the velocity. it is given by :
g is acceleration due to gravity
v = 6.26 m/s
So, the velocity with which the jumper strike the mat in the landing area is 6.26 m/s. Hence, this is the required solution.