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#1
golf bal hit from horizontal ground moves freely under gravity. horizontal component is 28m/s. vertical component is 14m/s. point from hit to wher it lands on ground is 80m. find in m/s the speed of the ball 2.5s after being hit.

M2 edexcel page 84 6c thats the question. is the answer in the back wrong?
0
16 years ago
#2
The answer is right. 2.5 seconds after being hit the horizontal component of the velocity does not change, it is still 14 m/s.

The vertical component changes, using v=u+at=14-9.8*2.5=-10.5 (the minus sign means that the ball is going downwards).

Therefore speed=sqrt[(10.5)^2+(28)^2]=30 m/s (2 s.f.)

|Darkness|
0
16 years ago
#3
I got the same answer as them.

First, find the displacement at 2.5 s:

x = ut + 1/2 at^2

x = -14 + 0.5 x 9.8 x 2.5^2 = -4.375 m

Then find the vertical component of the velocity:

v^2 = u^2 + 2ax

v = root[(-14)^2 + 2 x 9.8 x -4.375)] = 10.5 m/s.

Using Pythagoras:

v^2 = 10.5^2 + 28^2
v = root 894.25 = 29.9 m/s (30 m/s to 2 sf).
0
16 years ago
#4
Oh yeah, I could have used v = u + at. Lol.
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