Projectile Motion/Trajectories

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In summary, the person is seeking help with a problem and has found different answers from the answer key. They have explained their solution and are looking for someone to check their work. They are specifically trying to find the height of water jet after traveling horizontally for 2m.
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kobylorat
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Homework Statement



I've attached the problem. The answer key says the answer is (a), but I've reasoned it out to be (c). I'm not sure if the answer key is wrong, so can someone check my work? Thanks!

Homework Equations




The Attempt at a Solution



I've found that the initial vertical and horizontal speeds to be both 4√2 m/s using 8sin45 and 8cos45 respectively.
Then, dealing with only the vertical motion, I've found the time the water is in the air for by using the kinematic equation x = xi + vi*t + 0.5at^2.
Plugging values into the equation, I get: 1.5 = 4√2*t - 4.9t^2.
Solving for t, I get t=0.41269s and t=0.74177s. But, I don't use t=0.41269s because that is the time it takes for the ball to reach 1.5m as it is going up.
So, I take t=0.74177s and multiply it by the horizontal velocity, 4√2 m/s to obtain d=4.196084m, which means x is 2.19m.
 

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  • #2
kobylorat said:

Homework Statement



I've attached the problem. The answer key says the answer is (a), but I've reasoned it out to be (c). I'm not sure if the answer key is wrong, so can someone check my work? Thanks!

Homework Equations




The Attempt at a Solution



I've found that the initial vertical and horizontal speeds to be both 4√2 m/s using 8sin45 and 8cos45 respectively.
Then, dealing with only the vertical motion, I've found the time the water is in the air for by using the kinematic equation x = xi + vi*t + 0.5at^2.
Plugging values into the equation, I get: 1.5 = 4√2*t - 4.9t^2.
Solving for t, I get t=0.41269s and t=0.74177s. But, I don't use t=0.41269s because that is the time it takes for the ball to reach 1.5m as it is going up.
So, I take t=0.74177s and multiply it by the horizontal velocity, 4√2 m/s to obtain d=4.196084m, which means x is 2.19m.

You should find the height of water jet when it had traveled horizontally for 2m.
 

1. What is projectile motion?

Projectile motion is the motion of an object that is launched into the air and moves along a curved path under the influence of gravity. Examples of projectiles include a thrown baseball, a kicked soccer ball, or a fired bullet.

2. What factors affect the trajectory of a projectile?

The trajectory of a projectile is affected by the initial velocity, the angle of launch, the force of gravity, and air resistance. These factors determine the shape, height, and distance of the projectile's path.

3. How do you calculate the range of a projectile?

The range of a projectile can be calculated using the formula R = (V^2 * sin2θ)/g, where R is the range, V is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity. This formula assumes no air resistance.

4. Can the trajectory of a projectile be affected by external forces?

Yes, the trajectory of a projectile can be affected by external forces such as air resistance or wind. These forces can alter the shape, height, and distance of a projectile's path.

5. How is projectile motion used in real life?

Projectile motion is used in many real-life applications, such as sports, ballistics, and space exploration. Understanding the trajectory of a projectile can help in predicting the outcome of a baseball game, aiming a weapon, or launching a rocket into space.

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