# Few Questions

Gold Member
A string, 50cm long, has a stone of 100g tied to its end, and it is swung in a horizontal circle of radius 30cm. Calculate the tension in the string.

Because the string is longer than the radius of the circle, the path of the string must form a cone shape. Using the measurements, and forming a triangle, i found that the angle the string is forming is ~37 degrees to the vertical.

With this question, am i able to say that the vertical component of the tension force is the weight force from the stone? So ~0.98N? And then use that, along with the angle, to find the tension in the rope?

When you throw an orange up vertically from a moving ute, will it land back at the same spot from which it was thrown? explain.

For this question, i said if air resistance is neglected, and the use it travelling with uniform velocity, it will land on the same spot, since the orange will initially have the same horizontal velocity as the ute, and unless another force acts, they will both continute to have the same velocity.

A basketball player shoots for goal. The ball goes through the ring without touching it. If the ball is thrown from 50 degrees above the horizontal, and the basket is 2.5m in front of the player, and 1.3m higher than the point from which the ball was thrown, at what speed was the ball released?

I started by finding the time it takes for the ball to travel 2.5m in the horizontal plane.

$$\begin{array}{c} t = \frac{{s_h }}{{v_h }} \\ = \frac{{2.5}}{{v\cos 50}} \\ \end{array}$$

At that point in time, the vertical displacement should be 1.3m.
$$\begin{array}{c} s = ut + \frac{{at^2 }}{2} \\ 1.3 = \frac{{2.5v\sin 50}}{{v\cos 50}} - 4.9\left( {\frac{{2.5}}{{v\cos 50}}} \right)^2 \\ v = \pm 6.64 \\ \end{array}$$

So the initial velocity of the ball was 6.64 m/s at 50 degrees to the horizontal.

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These questions didnt come with answers, and im not really sure if ive done them correctly. If you could just look over them, and tell me if ive done anything wrong, that would be great.

Dan.

A string, 50cm long, has a stone of 100g tied to its end, and it is swung in a horizontal circle of radius 30cm. Calculate the tension in the string.

Because the string is longer than the radius of the circle, the path of the string must form a cone shape. Using the measurements, and forming a triangle, i found that the angle the string is forming is ~37 degrees to the vertical.

With this question, am i able to say that the vertical component of the tension force is the weight force from the stone? So ~0.98N? And then use that, along with the angle, to find the tension in the rope?

Yep, this seems correct to me. The vertical component of the tension force is the weight of the stone, the horizontal component is then the centripetal force of the circular motion.

When you throw an orange up vertically from a moving ute, will it land back at the same spot from which it was thrown? explain.

For this question, i said if air resistance is neglected, and the use it travelling with uniform velocity, it will land on the same spot, since the orange will initially have the same horizontal velocity as the ute, and unless another force acts, they will both continute to have the same velocity.

If the frame of reference is connected with the ute (the ute is travelling with uniform velocity and air resistance is neglected), it'll land on the same spot. However, if it is connected with the road, it won't land on the same spot - due to the initial horizontal velocity.

danago;1267728[b said:
A basketball player shoots for goal. The ball goes through the ring without touching it. If the ball is thrown from 50 degrees above the horizontal, and the basket is 2.5m in front of the player, and 1.3m higher than the point from which the ball was thrown, at what speed was the ball released?[/b]

I started by finding the time it takes for the ball to travel 2.5m in the horizontal plane.

$$\begin{array}{c} t = \frac{{s_h }}{{v_h }} \\ = \frac{{2.5}}{{v\cos 50}} \\ \end{array}$$

At that point in time, the vertical displacement should be 1.3m.
$$\begin{array}{c} s = ut + \frac{{at^2 }}{2} \\ 1.3 = \frac{{2.5v\sin 50}}{{v\cos 50}} - 4.9\left( {\frac{{2.5}}{{v\cos 50}}} \right)^2 \\ v = \pm 6.64 \\ \end{array}$$

So the initial velocity of the ball was 6.64 m/s at 50 degrees to the horizontal.

Just quickly flicking through your calcs on this one your ball does travel 2.5m in the horizontal but travels further than 1.3m in the vertical. This is because the ball has to come down through the top of the hoop.

Look at this as a ballistic trajectory question. The ball will travel an arc where the peak of the arc (max vertical motion) is less than 2.5m from the start point, the maximum range (max horizontal motion) is greater than 2.5m.

I haven't got my dynamics books to hand, what you need to do is work out the equation for a projectile at any point in the arc. possibly somebody who does have a dynamics book, or a large brain than mine can give you a pointer.

Gold Member
Yea i realise it travels further than 1.3m, but its displacement will still be 1.3m in the vertical plane wont it? The ring is 1.3m from the point of release, so wont the vertical displacement need to be 1.3m as it goes through?

Thanks for the help by the way everyone.