a long jumper leaves the ground at an angle of 30 degrees to the horizontal at a speed of 6 m/s. how far does he jump
+33653 horizontal velocity vh = 6*cos(30°) = 6*(√3)/2 = 3√3 m/s
initial vertical velocity vv = 6*sin(30°) = 6/2 = 3m/s
Using s = ut + at2/2 for change in vertical distance in time t, with acceleration a (-9.8m/s2) and initial velocity u (vv = 3m/s) we have
0 = 3*t - 9.8*t2/2 or t = 6/9.8 s (ignoring the t = 0 solution, which just represents staying still!).
The horizontal distance in time t is vh*t or 3√3*6/9.8 m
$${\frac{{\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{\mathtt{6}}}{{\mathtt{9.8}}}} = {\mathtt{3.181\: \!317\: \!809\: \!820\: \!386\: \!9}}$$
distance jumped ≈ 3.18m
+33653 horizontal velocity vh = 6*cos(30°) = 6*(√3)/2 = 3√3 m/s
initial vertical velocity vv = 6*sin(30°) = 6/2 = 3m/s
Using s = ut + at2/2 for change in vertical distance in time t, with acceleration a (-9.8m/s2) and initial velocity u (vv = 3m/s) we have
0 = 3*t - 9.8*t2/2 or t = 6/9.8 s (ignoring the t = 0 solution, which just represents staying still!).
The horizontal distance in time t is vh*t or 3√3*6/9.8 m
$${\frac{{\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{\mathtt{6}}}{{\mathtt{9.8}}}} = {\mathtt{3.181\: \!317\: \!809\: \!820\: \!386\: \!9}}$$
distance jumped ≈ 3.18m
