Question

The current world record for motorcycle jump is 77.0m, set by Jason Renie. Assume that he left the take off ramp at 12 degrees to the horizontal and that the take off and the landing heights are the same. Neglecting air drag determines his take off speed.

Answer 1

In the above question it is given to us that the motorcycle jumps i.e. it covers a distance along the horizontal of 77m such that he took off at an angle of 12 degrees with respect to the horizontal. As this jump is basically under the action of gravitational force and the motorcycle took off with some initial velocity, we can imply that the motion of the motorcycle will be similar to that of projectile motion. Hence we will use the expression for range in order to determine the initial speed of the motorcycle.

Formula used: $R=\dfrac{{{U}^{2}}Sin2\theta }{g}$

Complete step by step answer:
Let us say we project a body at an angle $\theta$ along the horizontal. If the initial velocity of the body is U than the distance covered by the body under acceleration due to gravity of Earth (g) i.e. range (R)is given by,
$R=\dfrac{{{U}^{2}}Sin2\theta }{g}m$
In the question we are asked to determine the initial speed of projection such that the distance covered by the motorcycle is 77m and makes an angle 12 degrees with the horizontal . Therefore from the above equation we get,
$\begin{aligned} & {{U}^{2}}=\dfrac{Rg}{Sin2\theta } \\\ & \Rightarrow {{U}^{2}}=\dfrac{77\times 9.8}{Sin2({{12}^{\text{o}}})}=\dfrac{754.6}{0.406}=1858.62 \\\ & \Rightarrow U=\sqrt{1858.62}m{{s}^{-1}} \\\ & \Rightarrow U=43.1m{{s}^{-1}} \\\ \end{aligned}$

Hence the motorcyclist must have taken off the ramp at an initial speed of 43.1 meters per second.

Note: It is to be noted that we have ignored the effect due to the air resistance on the bike. Therefore the above answer obtained will somewhat be similar to the actual speed of taking off. This is basically due to the fact that the effect in reality of air resistance of the bike is negligible due to its body.