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?

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.

To find the horizontal displacement of the melon on the ground, we need to consider the motion of the arrow. The vertical motion of the melon doesn't affect its horizontal displacement.

Let's break down the problem step by step:

1. First, we need to find the time it takes for the arrow to reach its highest point. To do this, we can use the vertical component of the arrow's initial velocity:

Viy = V * sin(theta)
Viy = 20 m/s * sin(60 deg)
Viy ≈ 17.32 m/s

We can use the equation for vertical motion to find the time at the highest point:

Vf = Viy - g * t
0 = 17.32 m/s - 9.81 m/s^2 * t

Solving for t:
t = 17.32 m/s / 9.81 m/s^2 ≈ 1.77 s

2. Now, we can find the horizontal component of the arrow's velocity:

Vix = V * cos(theta)
Vix = 20 m/s * cos(60 deg)
Vix ≈ 10 m/s

3. Using the horizontal velocity and the time at the highest point, we can find the horizontal displacement of the arrow:

Displacement = Vix * t
Displacement = 10 m/s * 1.77 s ≈ 17.7 m

4. Since the arrow hits the melon when it is at its highest point, and the melon falls straight down, the horizontal displacement of the melon is equal to the horizontal displacement of the arrow:

Horizontal displacement of the melon = Horizontal displacement of the arrow ≈ 17.7 m

Therefore, the horizontal displacement of the melon on the ground is approximately 17.7 meters.