Help Me! Solving Maths Basics - Car in Water

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SUMMARY

The discussion focuses on solving a physics problem involving projectile motion, specifically calculating the time it takes for a car to hit the water after leaving a quay 5 meters above water level and landing 6 meters away. The solution involves using the equations of motion, with gravitational acceleration set at 10 m/s². The time taken for the car to hit the water is determined to be 1 second, resulting in a vertical velocity of 5 m/s upon impact. Key concepts include the separation of horizontal and vertical motion in projectile dynamics.

PREREQUISITES
  • Understanding of basic physics concepts, particularly projectile motion.
  • Familiarity with equations of motion, including s = ut + 0.5at².
  • Knowledge of gravitational acceleration, specifically g = 10 m/s².
  • Ability to apply the Pythagorean theorem in velocity calculations.
NEXT STEPS
  • Study the principles of projectile motion in detail.
  • Learn how to apply the equations of motion to various scenarios.
  • Explore the concept of velocity vectors and their components.
  • Investigate real-world applications of projectile motion in engineering and physics.
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Students, educators, and anyone interested in understanding the fundamentals of physics, particularly those dealing with projectile motion and basic calculations in mechanics.

russs598
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I know I am thick!

I know this sound stupid but my kid has brought his homework home and although I am OK at higher maths i seem to have lost the ability to do the basics! :eek:
Can anyone help with this...

A car leaves a quay and ends in the water 6m away from the foot of the quay. The quay is 5m above water level.

Q. What is the time taken for the car to hit the water? Hence determine the velocity with which it hits the quay.

Cheers
 
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is there a time component to where the car splashed?
Its projectile motion. That you need to look up.

x=v_x0*t; x=6;
y=the acceleration formula with a=g...mmm can't remmeber but i think its
y+v_y0*t+g/2*t^2; y=5;

solve for v_x0, v_y0...and then do the pythagorean to get the speed. or if they want the velocity vector then your done.
 
The car falls vertically at the acceleration of gravity so the horizontal speed and distance of the car won't matter.

Assuming g=10m/s^2,
s = 5m, u = 0m/s, a = 10m/s^2, t = ?

s=ut + 0.5a(t^2)
t = sqrt (2s/a)
= 1s

therefore using v=s/t
v = 5m / 1s = 5m/s

Just keep in mind that projectiles have 2 separate speeds, the horizontal and the vertical and they don't mix up unless you want to find the resultant force.
 

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