Maximum Height and Range of a Projectile at a 60-Degree Angle

[azsx1]

Homework Statement

a cannon shoots a projectile @ ANGLE OF 60 DEGREES ABOVE HORIZONTAL W INITIAL SPEED OF 30M/S. CALCULATE MAX HT OF PROJECTILE AND ITS RANGE.

Homework Equations

U^2Sin^2O
30^2sin^260/9.8 = 67.5m not sure if this is correct!

The Attempt at a Solution

U^2Sin^2O
30^2sin^260/9.8 = 67.5m not sure if this is correct!

 

[tiny-tim]

Welcome to PF!

Hi azsx1! Welcome to PF!

(have a theta: θ and try using the X² tag just above the Reply box )

a cannon shoots a projectile @ ANGLE OF 60 DEGREES ABOVE HORIZONTAL W INITIAL SPEED OF 30M/S. CALCULATE MAX HT OF PROJECTILE AND ITS RANGE.
…
U^2Sin^2O
30^2sin^260/9.8 = 67.5m not sure if this is correct!

Yes (for the maximum height), except you're a factor of 2 out.

 

[azsx1]

Thank you! I think I am overthinking this, but I tried doing another way with d=v0t + .5at^2 formula. Thought for sure this would work! so i did and got a different answer! ugh. Can you help with range?

 

[tiny-tim]

Thank you! I think I am overthinking this, but I tried doing another way with d=v0t + .5at^2 formula.

Well, that'd never work, since you don't know d or t.

That's why you had to use a formula without t.

Can you help with range?

Yes … but you have to try it first.

 

[azsx1]

okay, i got t by imputting the initial velocity of the x/y components/ a for gravity. wouldn't this work? Also the range, i just plugged the numbers in formula d=v0t+.5at^2.

 

[tiny-tim]

Not following you.

Can you put the numbers in?​

 

[zgozvrm]

The time it takes the projectile to reach it's peak is calculated by:

$$t = \frac{v_{iy}}{g}$$

where [itex]v_{iy}[/tex] is the initial vertical velocity (the y-component) and [itex]g[/tex] is the acceleration of gravity (9.8 m/s/s)[/itex][/itex]