1) Since we know the y at the apex (1.41m) we know that vy= 0m/s so I used the e

Question

1) Since we know the y at the apex (1.41m) we know that vy= 0m/s so I used the equation vy^2=voy^2-2g(y-y0) and got voy=5.25m/s

2) Now I that I know voy I can switch over to the 40 degree angle part of this and find vox using the tangent of 40 degrees and I got vox=6.26m/s

3) Then I used R (total horizontal distance or range) = 2(vox)(voy)/g and I got x =6.70 m but this isn't right, what am I doing wrong?

7. [1pt] A small steel ball bearing with a mass of 25.0 g is on a short compressed spring. When aimed vertically and suddenly released, the spring sends the bearing to a height of 1.41 m. Calculate the horizontal distance the ball would travel if the same spring were aimed 40.0° from the horizontal

Explanation / Answer

Your Step 1) is correct.. Voy = 5.26 m/s

Your 2nd Step is wrong..

Here is how it should be :

Step 2)

Now in this case, Vo = 5.25

New Vox = 5.25*cos40 = 4.03 m/s

New Voy = 5.25*sin40 = 3.38 m/s

Now, time taken for flight(reach the ground) = 2*Voy/g = 2*3.38/9.8 = 0.6898 s

So, range, R = 2*voy*Vox/g = Vox*t = 4.03*0.6898 = 2.78 m <--------answer

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