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August 30, 2014

Posted by **Farah** on Sunday, December 5, 2010 at 5:18pm.

Ans: I tried the range formula - (Vi)^2* sin2theeta/(9.81), however i do not get the answer of 4.4m. i get 35.3m, where am i going wrong?

- Physics - check this work for horizontal displacem -
**MathMate**, Sunday, December 5, 2010 at 5:55pmThe range formula applies to an unobstructed projectile. In this case, the project has had an impact (with the water melon), and transfer of momentum has taken place, so the range formula does not apply.

You would have to calculate each step separately to find the final answer. The intermediate unknowns are:

- the horiz. velocity, v0, of the arrow

- the height of the water melon, h

- the combined horizontal velocity, V, after impact with the water melon

- the time, t, it takes for the water melon to drop from a height of h.

- Finally, the horizontal distance travelled during time t with the combined velocity of V.

Assume no air resistance.

Horizontal velocity,V0, of arrow

= v0*cos(θ)

= 20*cos(60°)

= 10 m/s

Vertical velocity of arrow, vy

= 20*sin(θ)

= 17.3 m/s

Highest point attained by arrow, h, can be solved from:

mgh = (1/2)m*vy²

h=m*vy²/mg = 15.3 m.

When the arrow hits the melon, the combined velocity can be obtained from conservation of momentum:

mv0=(m+M)V

or

V=mv0/(m+M)

=0.5*10/(0.5+1.5)=2.5 m/s

Time, t, for melon to fall to the ground

can be solved from

(1/2)gt^2 = h

The horizontal distance travelled is then

V*t

=2.5*t

I get 4.41 as distance.

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