A 1kg projectile is thrown North at an angle of (pi/4) from above the ground wit

Question

A 1kg projectile is thrown North at an angle of (pi/4) from above the ground with a starting speed of 2m/s. A wind blows towards the west with a force of 2 Newtons. Recall that the acceleration due to gravity is 9.81 towards the ground.

a) When does the projectile land?

b) Where does it land?

c) What is the max height of the projectile?

d) What are the vector equations for position, velocity, and acceleration? Be sure to describe your coordinate system (along which axis is N, W, and which axis points up).

Explanation / Answer

Everything North in positive, West is negative, east is positive

The velocity of the projectile in the East( positive x direction) is = 2cos 45 = 1.414 m/s
Velocity in the positive y direction( north) = 2 sin 45 = 1.414 m/s
acceleration in the x direction = force/ mass = 2/1 = - 1m/s^2 (west)
The final velocity of the projectile is 0 m/s(V) in the x direction
Using V = U + at
0 = 2 cos45 + (-1)t
Solving for t we get t = 1.414 seconds, this is when the projectile lands

using s = ut + 0.5at2
s = 1.414*1.414 + 0.5*(-1)(1.414)2 = 1 mt from the starting point

At the max height, the velocity in the y direction = 0= Vy
Using 2as = Vy2 - Uy2
2(-10)s = 0 - 1.4142
Solving for s we get s = 2/20 = 0.1 mts (max height)

All the equations of motions have been used above and the coordinate system has been described in the beginning and used through the problem
  

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