A Hollywood Daredevil plans to jump a canyon 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 69.6 m across. If he desires a a 3.2 second flight time
a. what is the correct angle of his launch ramp?
b.what is his correct launch speed?
c.what is the correct angle for the landing ramp?
d.what is his predicted landing velocity?
I answered this question last night. Have you changed screen names? The 60 m "original" information is irrelevant
yes, same person I just don't understand why im getting this problem wrong for a. predict the launch angle. I do tan inverse of 7.87/21.75 giving an angle of 19.9 degrees so I tried subtracting from 180 degrees but im not getting it right either way can you explain please?
consider the horizontal
69.6=speed*cosTheta*3.2
consider the vertical
-15=speed*sinTheta*3.2- 4.9 3.2^2
solve each side for speed first.
speed= 69.6/(3.2cosTheta)
speed= (-15+4.9*3.2^2)/sintheta*3.2
set them equal
69.6/(3.2cosTheta)=(-15+4.9*3.2^2)/sintheta*3.2
multipy both sides by sinTeta
69.6/3.2 * tanTheta= (-15/3.2 + 4.9*3.2)
solve for Theta.
tan theta= (-4.68+15.68)/21.75=.505
I get 26 deg. Check my work and thinking carefully. Very carefully.
ok thanks very much I understand the upward motion part of the problem. I am still have trouble with the downward motion can someone explain. Thanks for the help
To solve these problems, we'll need to use basic kinematic equations and principles of projectile motion. Let's break it down step by step:
a. To find the correct angle of the launch ramp, we can start by figuring out the time of flight. The horizontal distance traveled during a projectile motion is given by the equation: distance = velocity * time. In this case, the distance is 69.6 m, and the flight time is 3.2 seconds. Solving for the velocity, we get: velocity = distance / time.
velocity = 69.6 m / 3.2 s = 21.75 m/s
Now, let's focus on the vertical component of the motion. When the daredevil jumps a canyon, it can be assumed that he's launching from the same height as the landing ramp. Given that there is a 15 m drop, the launch height is 15 m.
Using the kinematic equation: height = (initial velocity * sin(angle))^2 / (2 * gravity), we can find the correct angle. Rearranging the equation and plugging in the values:
sin(angle) = sqrt((2 * gravity * height) / (velocity^2))
angle = arcsin(sqrt((2 * gravity * height) / (velocity^2)))
b. To find the correct launch speed, we already found the velocity in part a. The launch speed is simply the magnitude of the velocity vector, so we can take the absolute value:
launch speed = |velocity| = |21.75 m/s| = 21.75 m/s
c. Now, let's calculate the correct angle for the landing ramp. We know that the flight time is 3.2 seconds and the horizontal distance is 69.6 m. We can use this information to find the landing speed (vertical component) using the equation: distance = 0.5 * gravity * (time^2) - (initial velocity * sin(angle)) * time.
Solving for the landing speed:
69.6 m = 0.5 * gravity * (3.2 s)^2 - launch speed * sin(angle) * 3.2 s
landing speed = (69.6 m + 21.75 m/s * sin(angle) * 3.2 s) / (0.5 * gravity * (3.2 s)^2)
Once we find the landing speed, we can use a similar kinematic equation (height = velocity^2 * sin^2(angle) / (2 * gravity)) to calculate the correct angle:
sin(angle) = sqrt((2 * gravity * height) / (landing speed^2))
angle = arcsin(sqrt((2 * gravity * height) / (landing speed^2)))
d. The predicted landing velocity can be found using another kinematic equation: velocity = sqrt((horizontal speed)^2 + (vertical speed)^2). We already have the landing speed from part c, and the horizontal speed is the horizontal distance (69.6 m) divided by the flight time (3.2 s):
horizontal speed = distance / time = 69.6 m / 3.2 s = 21.75 m/s
Now we can calculate the predicted landing velocity:
landing velocity = sqrt((21.75 m/s)^2 + landing speed^2)
Remember to use the value of acceleration due to gravity (approximately 9.8 m/s^2) in the calculations.