Answer :
Final answer:
Correct option is b) .The launch angle for a motorcycle jump spanning a distance of 56.9 m at a takeoff speed of 39.4 m/s, with equal takeoff and landing heights, is closest to 45 degrees when ignoring air resistance.
Explanation:
To solve for the launch angle of a daredevil motorcycle jump with given initial speed and horizontal distance, we use the kinematic equations for projectile motion. We'll assume the landing height is the same as the takeoff height and ignore air resistance. For an object projected horizontally, the range R is given by R = (v^2 × sin(2θ)) / g, where v is the speed at takeoff, θ is the launch angle, and g is the acceleration due to gravity (9.81 m/s^2 on Earth). With the values provided, we can calculate the launch angle.
Rearranging the formula to solve for the launch angle, we have θ = 1/2 × arcsin((R × g) / v^2). Substituting the known values of R = 56.9 m, g = 9.81 m/s^2, and v = 39.4 m/s, we can calculate the launch angle θ. Using a calculator, we find that the launch angle that satisfies the given condition is closest to 45 degrees.