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Impact speed

  • Thread starter Jjolly65
  • Start date
  • #1
3
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Homework Statement



A ball is thrown toward a cliff of height h with a speed of 23.88 m/s and an angle of 52° above horizontal. It lands on the edge of the cliff 3.49 s later.


What is thy ball impact speed?


2. The attempt at a solution

shouldnt it be
Vx= Vo*cos(theta)
Where Vx= 23.88*cos(52)= 14.70 m/s?
Thats what im getting but its telling me I'm wrong... Am I using the wrong equation?
 

Answers and Replies

  • #2
161
0
You have calculated the horizontal component of the velocity.

But the impact speed will be the magnitude of the resultant of the horizontal and vertical components of velocity after 3.49s.

The horizontal velocity doesn't change (assuming no air resistance), but you need to work on the vetical component.

Have you done much projectile motion / motion in 2 dimensions work before?
 
  • #3
3
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How do I find the vertical component then?

I found the hieght of the cliff to be 5.99m and the max height of the ball in the air would be 18m... I don't know if that helps...
 
  • #4
161
0
The initial vertical component of the velocity is given by Uy = Vo*sin(theta).

You don't need any of that other business.

You can use Vy = Uy + Ay*t with the given time to find the vertical component of the velocity.

Remember to define a positive and negative. It will probably be easiest to say that your acceleration is negative.
 

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