Parabolic Motion, 45 degrees

[RobSchneider]

Homework Statement

A cricket ball is thrown upwards at an angle of 45 degrees and pitches 20m from the thrower. What was the balls initial speed and how high does it rise?

Homework Equations

The Attempt at a Solution

Initial Speed- 19.7 m/s
Height- 10(??) m

 

[PeterO]

Homework Statement

A cricket ball is thrown upwards at an angle of 45 degrees and pitches 20m from the thrower. What was the balls initial speed and how high does it rise?

Homework Equations

The Attempt at a Solution

Initial Speed- 19.7 m/s
Height- 10(??) m

Maximum height can be solved with logic.

When projected at 45 degrees, the initial vertical and horizontal components of velocity are equal.

The horizontal velocity remains the same throughout the flight.
The vertical component reduced to zero in the first half of the trip, then gains the same magnitude of velocity in the second half [coming down rather than going up.

Consider the gain of maximum height.
Velocity drops from the initial vel to zero at a steady rate, so the average velocity is one half of that.
The horizontal velocity has remained the same throughout.
That means the average horizontal velocity is twice the average vertical velocity.
That means the object will travel twice as far horizontally as it does vertically.
All this takes place during the first half of the flight.
in the whole flight, the ball traveled 20m horizontally, so in the first half of the trip, the ball traveled only 10m - and gained a height only half of that.

 

[trolling]

An equation that might be useful in this problem is the range equation, where ∆x = ((v-init.)^2 * sin (2α)/g), where α is the angle & g is gravitational acceleration (9.8 m/s^2). Solve for v-init to get 14 m/s.

You then divide the initial speed into its components by dividing both of them (x & y) by sqrt(2), since the angle is 45 degrees. Use the y-component as the initial velocity & 0 as the final velocity in the equation ((v-final)^2) = ((v-init)^2) - 2*g*y, where g is the gravitational acceleration & y is the change in height.
Solve for y to get 5 m.

 

[PeterO]

Homework Statement

A cricket ball is thrown upwards at an angle of 45 degrees and pitches 20m from the thrower. What was the balls initial speed and how high does it rise?

Homework Equations

The Attempt at a Solution

Initial Speed- 19.7 m/s
Height- 10(??) m

As you will see by the thought solution, and the formula solution, you Height answer is incorrect. The Initial Speed figure is also incorrect.
Perhaps if you showed how you got those figures we may be able to see where you went wrong.

 

[Nickie]

I would first like to clarify that the term "initial speed" is commonly known as the initial velocity, which is the speed and direction at which an object is launched. With that being said, let us approach this problem using the equations of parabolic motion.

The given information tells us that the ball is thrown upwards at an angle of 45 degrees and pitches (or lands) 20m away from the thrower. This means that the ball follows a parabolic path, where it has both horizontal and vertical components of motion. The initial velocity of the ball can be broken down into its horizontal and vertical components as follows:

Horizontal component: v0x = v0cosθ
Vertical component: v0y = v0sinθ

Where v0 is the initial velocity, θ is the angle of launch (45 degrees in this case), v0x is the horizontal component of the initial velocity, and v0y is the vertical component of the initial velocity.

Now, using the given information, we can solve for the initial velocity (v0) as follows:

20m = v0x * t
Where t is the time it takes for the ball to travel 20m horizontally. Since we do not have the value of time, we will need to find it using the vertical component of motion. We know that the ball is thrown upwards, reaches its maximum height, and then falls back down. At the highest point, the vertical velocity (vy) is 0, and the time taken to reach that point is half of the total time of flight (t/2). Therefore, we can use the following equation to find the time of flight (t):

0 = v0y + gt/2
Where g is the acceleration due to gravity (9.8 m/s^2). Solving for t, we get t = 2v0y/g.

Now, substituting this value of t in the equation for horizontal motion, we get:

20m = v0cosθ * 2v0y/g
Simplifying, we get v0 = 19.7 m/s.

Therefore, the initial velocity of the ball is 19.7 m/s, which is the speed at which it was thrown upwards at an angle of 45 degrees.

To find the maximum height reached by the ball, we can use the equation for vertical motion:

h = v0yt - 1/2