A Hollywood daredevil plans to jump the canyon shown in the figure on a motorcycle. There is a 15. m drop and the horizontal distance originally planned was 60. m but it turns out the canyon is really 65.7 m across. If he desires a 3.2-second flight time, (a) what is the correct angle for his launch ramp (deg)? (b) What is his correct launch speed? (c) What is the correct angle for his landing ramp (give a positive angle below the horizontal)? (d) What is his predicted landing velocity. (Neglect air resistance.)
For (a) I got 28.2 degrees already
To solve this problem, we can use the principles of projectile motion. Let's break down the question into different parts and find the solutions step by step:
(a) Correct angle for the launch ramp (deg):
To find the correct angle for the launch ramp, we can use the equation for the range of a projectile, which is given by:
Range = (Initial velocity^2 * sin(2*theta)) / g
where theta is the angle of the launch ramp and g is the acceleration due to gravity.
Given that the range is 65.7 m and the acceleration due to gravity is approximately 9.8 m/s^2, we can rearrange the equation to find theta:
theta = (1/2) * arcsin((Range * g) / (Initial velocity^2))
Since the flight time is equal to 3.2 seconds, we can express the initial velocity in terms of the range and flight time:
Initial velocity = Range / (cos(theta) * flight time)
Substituting this value back into the expression for theta, we can solve for the correct angle.
(b) Correct launch speed:
To find the correct launch speed, we can use the derived formula for initial velocity as mentioned above:
Initial velocity = Range / (cos(theta) * flight time)
Substitute the correct angle value obtained in part (a) and the given range and flight time to get the correct launch speed.
(c) Correct angle for the landing ramp:
To find the correct angle for the landing ramp, we can consider the vertical displacement during flight. The horizontal distance has remained the same, so the horizontal component of the landing velocity will be the same as the launch velocity.
The vertical component of the landing velocity can be found by considering the vertical displacement in free-fall motion:
Vertical displacement = Initial velocity * sin(theta) * flight time - (1/2) * g * (flight time^2)
Given that the vertical displacement is equal to a 15 m drop (negative), solve for the correct angle below the horizontal that satisfies this condition.
(d) Predicted landing velocity:
Since we know the horizontal component of the landing velocity is the same as the launch velocity, we can use trigonometry to find the magnitude of the landing velocity.
Landing velocity = sqrt((Horizontal velocity)^2 + (Vertical velocity)^2)
Substitute the correct launch speed obtained in part (b) and the vertical component of the landing velocity obtained in part (c) to find the predicted landing velocity.
By following these steps, you can find the correct angle for the launch ramp, the correct launch speed, the correct angle for the landing ramp, and the predicted landing velocity.