2D Kinematics - Projectile Motion Question

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SUMMARY

The projectile motion problem involves a daredevil shot from a cannon at an angle of 33.7° with an initial speed of 25.5 m/s, aiming to determine the height of a net positioned 43.7 m away. The initial horizontal velocity (vox) is calculated as 21.1 m/s, and the initial vertical velocity (voy) is 14.1 m/s. Using these values, the time of flight can be determined, which is essential for calculating the vertical displacement (dy) needed to find the height of the net above the cannon's mouth. The acceleration due to gravity is 9.81 m/s², which influences the vertical motion of the projectile.

PREREQUISITES
  • Understanding of 2D kinematics principles
  • Knowledge of trigonometric functions (sine and cosine)
  • Familiarity with projectile motion equations
  • Basic algebra for solving equations
NEXT STEPS
  • Calculate the time of flight using the formula: time = horizontal distance / horizontal velocity
  • Use the time of flight to find the vertical displacement using the equation: dy = voy * time + 0.5 * a * time²
  • Explore the effects of varying launch angles on projectile motion
  • Investigate the impact of different initial speeds on the height of the projectile
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Students studying physics, particularly those focusing on kinematics and projectile motion, as well as educators seeking to enhance their teaching methods in these topics.

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Homework Statement



A daredevil is shot out of a cannon at 33.7◦ to the horizontal with an initial speed of 25.5 m/s. A net is positioned at a horizontal distance of 43.7 m from the cannon from which the daredevil is shot.
The acceleration of gravity is 9.81 m/s^2
At what height above the cannon’s mouth should the net be placed in order to catch the daredevil?
Answer in units of m

Given: θ = 33.7; vo = 25.5 m/s; dx = 43.7m; a = 9.81m/s^2
Find: dy

Homework Equations



vox = vo(cosθ)
voy = vo(sinθ)

[I'm stuck on what other equations I can use]


The Attempt at a Solution



I first drew a diagram with no problem.
vox = (25.5m/s)cos33.7 = 21.1m/s @ a = 0m/s^2
voy = (25.5m/2)sin33.7 = 14.1m/s @ a = -9.81m/s^2

So we now know initial velocity in the x direction is 21.1m/s, and initial velocity in the y direction is 14.1m/s. I am a little confused on where to go from here.
 
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You know how far the shot goes in the x direction (43.7 m) and the initial x component of velocity, so you can calculate the time of flight.
 

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