Understanding Projectiles (Example Problem Given)

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In summary, the golfer is seeking help with tackling projectile problems and wants to know the initial velocity of a ball that needs to travel 130M horizontally and 20M vertically, with a tee-off point 10M above the landing spot. The 3 kinematic formulas are used: S= VoT + 1/2at², V=Vo + at, and V²=Vo² + 2as. A diagram is usually drawn and the problem is broken down into horizontal and vertical components. The equations used are d = vt for horizontal and V = Vi + at and d = Vi*t + .5*a*t² for vertical. There can be multiple solutions, depending on the angle at which the ball is
  • #1
smashbrohamme
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I don't know how to tackle projectiles and would like to have someone sort of show me what you look for in projectiles and how you normally tackle projectile problems.

The golfer wants to hit the ball the same horizontal distance of 130M and 20M above his tee-off point as in Problem 10-32, but in this case, the tee-off point is 10M above the landing spot of the ball. What is the ball's initial velocity.

*10-32 just states the same problem but doesn't have a 10M tee-off point*


Im using the 3 kinematic formulas.

S= VoT + 1/2at²
V=Vo + at
V²=Vo² + 2as

I just don't know where to begin other then to draw a diagram. I understand somewhere in this problem I must use trig functions cos, sin but I am seriously stuck here..How do you guys normally tackle these projectile problems...break it down like I am a 1st grader!
 
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I have yet to see one that doesn't yield to the basic high school approach.
Make two headings for HORIZONTAL and VERTICAL. Under horizontal, write d = vt. Under vertical write V = Vi + at and d = Vi*t + .5*a*t².
Usually you only use two of the equations, but you don't know in advance which two, so you put the knowns into all three and then take a look. Sometimes you have to solve two of them as a system of equations.

In this particular example, there appear to be an infinite number of solutions. If you shoot at 45 degrees, there is a speed v1 that works. If you shoot a little lower or higher, a slightly larger speed is needed. And so on.
 

Related to Understanding Projectiles (Example Problem Given)

1. What is a projectile?

A projectile is an object that is launched or thrown and moves through the air under the influence of gravity, without any additional propulsion. Examples of projectiles include a baseball being thrown, a rocket being launched, or a bullet being fired from a gun.

2. How is the trajectory of a projectile determined?

The trajectory of a projectile is determined by its initial velocity, angle of launch, and the force of gravity acting on it. These factors can be calculated using mathematical equations, such as the projectile motion equations, to determine the path of the projectile.

3. What is the difference between range and maximum height in projectile motion?

The range of a projectile is the horizontal distance it travels before hitting the ground, while the maximum height is the highest point the projectile reaches in its trajectory. Both of these quantities are affected by the initial velocity and angle of launch of the projectile.

4. How does air resistance affect projectile motion?

Air resistance, also known as drag, can affect the trajectory of a projectile by slowing it down. This is because air resistance creates a force in the opposite direction of the projectile's motion, causing it to lose velocity and deviate from its expected path. The effect of air resistance becomes more significant at higher velocities and for objects with a larger surface area.

5. Can you provide an example problem of projectile motion and how to solve it?

Example Problem: A ball is thrown from a height of 5 meters with an initial velocity of 10 m/s at an angle of 45 degrees. What is the range and maximum height of the ball?To solve this problem, we can use the projectile motion equations. First, we can calculate the time the ball spends in the air using the equation t = 2v₀sinθ/g, where v₀ is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity. Plugging in the values, we get t = 2(10 m/s)sin(45°)/9.8 m/s² = 1.02 seconds.Next, we can use the equations x = v₀cosθt and y = v₀sinθt - 1/2gt² to calculate the range and maximum height. Plugging in the values, we get x = (10 m/s)(cos 45°)(1.02 s) = 7.07 meters for the range, and y = (10 m/s)(sin 45°)(1.02 s) - 1/2(9.8 m/s²)(1.02 s)² = 2.04 meters for the maximum height.Therefore, the range of the ball is 7.07 meters and the maximum height is 2.04 meters.

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