The footballer kicked the ball and it was caught by the opposing goalkeeper, 4 metres above the ground. The ball was traveling at 10 m/s^2.Calculate the kinetic energy of the ball just before it was caught.
Solution.
"v_0=10m\/s;"
"h=4m;"
"W_k-?;"
"\\dfrac{mv_0^2}{2}=mgh_{max}\\implies h_{max}=\\dfrac{v_0^2}{2g};"
"h_{max}=\\dfrac{100}{2\\sdot9.81}=5.1m;"
"mgh_{max}=mgh+\\dfrac{mv^2}{2}\\implies W_k= \\dfrac{mv^2}{2}=mgh_{max}-mgh;"
"W_k=m(gh_{max}-gh);"
"W_k=m(9.81\\sdot5.1-9.81\\sdot4)=10.79mJ;"
Answer: "W_k=10.79m J."
Comments
Leave a comment