Problema Solution
An athlete in a high jump competition jumps at angle of 45 degrees and just crosses over a horizontal bar 2.5m above the ground. What is the initial speed of the athlete?
Answer provided by our tutors
We assume the initial speed is V.
The time t = 2*V*sin45/g
The horizontal distance S = V*cos45*t = V*cos45*2*V*sin45/g = V^2/g
So
V^2/g = 2.5
V = 4.95 m/s
So the initial speed is 4.95 m/s