How Do You Calculate Initial Speed and Maximum Height in Parabolic Motion?

[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.