Projectile Motion Question 1 of Grade 12 Physics (University Preparation, SPH4U) of Ontario Curriculum, Canada

Question

Khoi thinks he can throw the rock further, so he spins the same 2kg rock on a 0.75m rope at an angle of

30° from the horizontal. Assuming that Khoi lets go of the rock when it was 2.15m above the ground

and the tension in the rope was 150N when he let go, how far did Khoi’s rock go before hitting

the ground. If you remember projectile motion you can answer this one.

Answer

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Legend of Figure 1:

G = 2g (N) gravity

g acceleration of gravity

OH horizontal line

V rock speed at point O

O start point when rock was let go (the same point as point A of figure 2)

T rope tension when rock was let go

F centripetal force when rock was let go

α angle between rope tension T and horizontal line OH

30° angle between rock speed V and horizontal line OH

The angle between force F and line OH equals 90° - 30° = 60°. It’s easy to get the following equation

system from the forces relationship in Figure 1.

From this equation system and the following triangle formula, 

From the formula

,
with m being rock’s mass 2kg, and r being rope’s length 0.75m, we get

Legend of Figure 2:

V rock speed when being let go

Vv vertical projection of rock speed when rock was let go

Vh horizontal projection of rock speed when rock was let go

30° angle between rock speed V and horizontal line OH

A start point when rock was let go

O point A’s horizontal projection on the ground

P parabola vertex of rock’s motion track

Q intersection of horizontal line AQ and rock motion track

R point where rock landed ground

D distance of points O and R

Vv = V.sin30° = 3.953.

Since Vv = g.t1, with t1 being the time when the rock moved from points A to P, we have

.

t2 is the time when the rock moved from points Q to R. From the equation

,
with 2.15 being the distance of points O and A, we get t2 = 0.3722.

The total time when the rock moved from points A to R is tT = 2 t1 + t2 = 1.179

D = Vh . tT = V . cos30° . tT = 8.08

The distance of points A and R is

.
Above is the calculation considering the gravity effect on the centripetal force F. If we ignore the effect, we have F = 150N, V = 7.5, t1 = 0.383, h = 0.7188, t2 = 0.382, tT = 1.148, D = 7.46, and the distance of points A and R = 7.76 ≈ 7.8m.


                                           


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