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

A 1.5kg melon is placed on top of the post. A girl lying on the ground fires a 500g arrow at ground level with velocity of 20m/s, 60deg above the horizontal, and hits the melon when the arrow is at its highest point. The melon, with the arrow embedded, falls to the ground. Find the horizontal displacement of the melon on the ground.

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 5:55pm
The 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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