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Margarita [4]
3 years ago
6

One strategy in a snowball at a high angle over level ground. While your opponent is watching this first snowball, you throw a s

econd one at low angle times to arrive before or at the same time as the first one . Assume both snowballs are throw with a speed of 33.2 m/s. The first o e is thrown at an angle of 57 degrees with respect to the horizontal. At what angle should the second snowball be throw to arrive at the same point as the first? Answer in units of degrees
Physics
1 answer:
siniylev [52]3 years ago
8 0

Answer:

33°

Explanation:

We are given;

Speed at which both snowballs are thrown; v = 33.2 m/s

Angle at which snowballs are thrown with respect to the horizontal; θ = 57°

Now,we want to find out the angle at which the second snowball should be thrown in order to arrive at the same point as the first.

To calculate this angle, we will use complementary angle concept.

Now, because the target is in the same place, there will be two launch angles that will make the snow ball to be placed on the target.

The is calculated from;θ1 = 45° - (57° - 45°) = 33°

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Answer:

Time to pass the train=0.05 h

How far the car traveled in this time=4.75 Km

Explanation:

We have that the train and the car are moving in the same direction, the difference between the speed of the vehicles is:

\Delta V=V_{car}-V_{train}=95km/h-75km/h=20km/h

We will use this difference in the speed of the car an train to calculate how much time take the car to pass the train. For this we have that the train is 1km long and the car is moving with a speed of 20km/h (we use this value because is the speed that the car have in advantage of the train) then for a movement with a constant speed we have:

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Where x is the distance, t is the time and v is the speed. using the data that we have:

V=\dfrac{x}{t}=\dfrac{1km}{20km/h}=0.05h

This is the time that the car take to pass the train. Now to calculate how far the car have traveled in this time we have to considered the speed of 95Km/h of the car, then:

V=\dfrac{x}{t}\\x=v\cdot t\\x=95km/h\cdot 0.05h\\x=4.75km

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