2D Kinematics - Projectile Motion Question

Click For Summary
To determine the height at which the net should be placed to catch the daredevil, the initial velocities in both the x and y directions have been calculated as 21.1 m/s and 14.1 m/s, respectively. The horizontal distance to the net is 43.7 m, allowing for the calculation of the time of flight using the formula time = distance/velocity. Once the time of flight is established, the vertical displacement can be found using the kinematic equation for vertical motion, incorporating the acceleration due to gravity. The solution requires applying these principles to find the height above the cannon's mouth where the net should be positioned. This approach will yield the necessary height in meters.
5minutes
Messages
3
Reaction score
0

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.
 
Physics news on Phys.org
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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
16
Views
3K
Replies
18
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
33
Views
4K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K