Projectile Motion - Initial Velocity

[SteveJW]

Good Afternoon, wondering if I could get some assistance. I'm not formally learning physics its just an interest I have, and try to complete problems during my spare time.

I have one particuar question that I'm stumped with, so any help would be great!

Q: You are asked by a film director to organise a stunt scene in which a motorbike mounts a ramp before safely jumping a wall. The wall is 4m high and the end of the ramp is 0.5m high and placed 10m back from the wall. What is the speed the stunt rider must drive off the ramp in order to jump the wall?

I understand that I need to split into horizontal and vertical components(?) then use uniform motion equation to get time taken and ultimately the initial velocity? Is this correct?

Kind regards

Steve

 

[ashishsinghal]

what is the angle of the ramp

 

[SteveJW]

it doesn't say, but would I be correct in saying that Tan<x>3.5/10?

 

[SteveJW]

Sorry, should have said...tanx=3.5/10.

As I say l, any help is appreciated

 

[rwisz]

Hello Steve,

Thank you for reaching out. It's great that you have an interest in physics and are trying to solve problems on your own. I'd be happy to assist you with your question on projectile motion.

First, you are correct in your approach of splitting the motion into horizontal and vertical components. This is known as vector decomposition and it is a fundamental concept in projectile motion.

To solve this problem, we can use the equations of motion for projectile motion. The initial velocity (u) of the motorbike can be found by using the equation:

u = √(2gh)

Where g is the acceleration due to gravity (9.8 m/s²) and h is the height of the ramp (0.5 m). Plugging in these values, we get:

u = √(2*9.8*0.5) = 3.13 m/s

This is the initial velocity of the motorbike as it leaves the ramp. Now, we can use the range equation to find the speed (v) at which the motorbike must leave the ramp in order to clear the 4m high wall. The range equation is given by:

R = (v²sin2θ)/g

Where R is the range (10 m in this case), θ is the angle of projection (which we can assume to be 45° for maximum range), and g is the acceleration due to gravity. Plugging in the values, we get:

10 = (v²sin90)/9.8

Solving for v, we get:

v = √(10*9.8) = 9.9 m/s

Therefore, the speed at which the stunt rider must drive off the ramp is 9.9 m/s.

I hope this helps you understand the concept of projectile motion and how to apply it to solve problems. Keep up the good work and don't hesitate to reach out if you have any further questions.

Best regards,