a physics professor did daredevil stunts in his spare time. his last stunt was to attempt to jump across a river on a motorcyle. the take-off ramp was inclined at 53°, the river was 40m while, and the far bank was 15m lower than the top of the ramp. the river itself was 100m below ramp. what should his velocity have been at the top of the ramp to have just made it to the edge of the far bank?
+33653 Chris, I did mean 53° not 56°. Goodness knows where the 56 came from, there isn't a 6 anywhere in the question! Just put it down to early onset senility!
$${\mathtt{v}} = {\sqrt{{\frac{\left({\frac{{\mathtt{9.8}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{40}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{53}}^\circ\right)}}}\right)}^{{\mathtt{2}}}}{{\mathtt{2}}}}\right)}{\left({\mathtt{40}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{53}}^\circ\right)}{\mathtt{\,\small\textbf+\,}}{\mathtt{15}}\right)}}}} \Rightarrow {\mathtt{v}} = {\mathtt{17.831\: \!156\: \!614\: \!371\: \!128}}$$
v ≈ 17.8m/s
+33653 Ignoring air resistance, horizontal velocity is constant at v*cos(56°). Time is distance over velocity, so time to cross is t = 40/(v*cos(56°).
Change in vertical distance is given by -15 = v*sin(56°)*t - 9.8*t2/2 or
-15 = v*sin(56°)*40/(v*cos(56°)) - 9.8*(40/(v*cos(56°)))2/2
-15 = 40*tan(56°) - 9.8*(40/(v*cos(56°)))2/2
(9.8*(40/cos(56°))2/2)/(40*tan(56°) + 15) = v2
$${\mathtt{v}} = {\sqrt{{\frac{\left({\frac{{\mathtt{9.8}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{40}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{56}}^\circ\right)}}}\right)}^{{\mathtt{2}}}}{{\mathtt{2}}}}\right)}{\left({\mathtt{40}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{56}}^\circ\right)}{\mathtt{\,\small\textbf+\,}}{\mathtt{15}}\right)}}}} \Rightarrow {\mathtt{v}} = {\mathtt{18.369\: \!394\: \!701\: \!106\: \!074}}$$
or v ≈ 18.4m/s
+129840 Alan....where did the "56°" come from???
Did you mean "53°" instead ???
+33653 Chris, I did mean 53° not 56°. Goodness knows where the 56 came from, there isn't a 6 anywhere in the question! Just put it down to early onset senility!
$${\mathtt{v}} = {\sqrt{{\frac{\left({\frac{{\mathtt{9.8}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{40}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{53}}^\circ\right)}}}\right)}^{{\mathtt{2}}}}{{\mathtt{2}}}}\right)}{\left({\mathtt{40}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{53}}^\circ\right)}{\mathtt{\,\small\textbf+\,}}{\mathtt{15}}\right)}}}} \Rightarrow {\mathtt{v}} = {\mathtt{17.831\: \!156\: \!614\: \!371\: \!128}}$$
v ≈ 17.8m/s
