Projectiles launched at an angle question

In summary: The height is all about the y component of the motion, so you have v and angle A.The y component of the initial velocity is v.sinA ... well done.The acceleration due to gravity in the +y direction is -g (no need to put numbers in yet)t== vf - vi / a , t=-9.07980 - 9.07980 / -9.81m/s^2 At what y-position is the speed a maximum?At what y-position is the speed a minimum?While it is going up, what happens to it's vertical speed?While it is going down, what happens to it's vertical speed?
  • #1
SmallPub
19
0

Homework Statement

An object is thrown from the ground into the air with a velocity of 20.0m/s at an angle of 27.0° to the horizontal what is the maximum height?

v= 20.0 m/s a=-9.81m/s^2 angle = 27.0

Homework Equations

a=v/t , v= d/t

The Attempt at a Solution

sin(27.0°) = y/20.0m/s (first i solved the vertical direction)

y = sin(27.0°) x 20.0m/s

y =9.07980 m/s

a=v/t (times t then divided by a)

t== vf - vi / a , t=-9.07980 - 9.07980 / -9.81m/s^2

t=1.8511s

cos(27.0) = x/20.0m/s

x = cos(27.0) x 20.0m/s

x= 17.8201m/s

v=d/t , d= VxT

d = 17.8201m/s x 1.8511s

d = 30.0 m ?

I thought you solved it the same way as you solve the object going horizontally. I am really confused how to solve for height.

Thanks
 
Physics news on Phys.org
  • #2
I thought you solved it the same way as you solve the object going horizontally. I am really confused how to solve for height.
The height is not the same as the range - therefore it is not the same method.
You have to think about why you are doing each step and not just memorize the steps.
The height is all about the y component of the motion ... so you have v and angle A.

The y component of the initial velocity is v.sinA ... well done.
The acceleration due to gravity in the +y direction is -g (no need to put numbers in yet)

t== vf - vi / a , t=-9.07980 - 9.07980 / -9.81m/s^2
... what is the y-velocity when the projectile has reached it's maximum height?

Do you know a kinematic equation which will give you the final displacement, given the initial and final velocities, and the acceleration?
 
  • #3
Simon Bridge said:
... what is the y-velocity when the projectile has reached it's maximum height?

would it be 9.0798m/s?

Do you know a kinematic equation which will give you the final displacement, given the initial and final velocities, and the acceleration?

d=vi.t + 1/2.a.t^2 ?
 
  • #4
Simon Bridge said:
... what is the y-velocity when the projectile has reached it's maximum height?

SmallPub said:
would it be 9.0798m/s?

If that were true, why isn't the projectile still traveling upward?

If something is not moving in a certain direction, what is its velocity in that direction?
 
  • #5
would [ y-velocity when the projectile has reached it's maximum height] be 9.0798m/s?
What is the value of y when the ball has that speed??

You need to watch a projectile in motion. Get something you can throw - a ball is good.
Throw it straight up. Catch it. Watch carefully - do it several times.

Observe:
At what y-position is the speed a maximum?
At what y-position is the speed a minimum?
While it is going up, what happens to it's vertical speed?
While it is going down, what happens to it's vertical speed?
When it is at exactly the top of it's trajectory, what is it's speed?

d=vi.t + 1/2.a.t^2 ?
That's a good kinematic equation - but it has time in it.
You don't know the time ... you need one for displacement, given the initial and final velocities and the acceleration.
i.e. you want d, given vi, vf, and a.
 

1. What is the formula for calculating the range of a projectile launched at an angle?

The formula for calculating the range of a projectile launched at an angle is R = (v^2sin2θ)/g, where R is the range, v is the initial velocity of the projectile, θ is the launch angle, and g is the acceleration due to gravity.

2. How does the launch angle affect the range of a projectile?

The launch angle directly affects the range of a projectile. The optimal launch angle for maximum range is 45 degrees, as it allows for equal vertical and horizontal displacement.

3. What factors can affect the trajectory of a projectile launched at an angle?

The trajectory of a projectile launched at an angle can be affected by the initial velocity, launch angle, air resistance, and the force of gravity.

4. How can we calculate the maximum height of a projectile launched at an angle?

The maximum height of a projectile launched at an angle can be calculated using the formula h = (v^2sin^2θ)/2g, where h is the maximum height, v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

5. What is the difference between the range and the maximum height of a projectile launched at an angle?

The range is the horizontal distance covered by a projectile, while the maximum height is the highest point reached by the projectile. The range is affected by the initial velocity and launch angle, while the maximum height is affected by the initial velocity and launch angle, as well as the force of gravity.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
977
  • Introductory Physics Homework Help
2
Replies
38
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
2
Replies
36
Views
2K
  • Introductory Physics Homework Help
Replies
19
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
798
  • Introductory Physics Homework Help
Replies
11
Views
668
  • Introductory Physics Homework Help
Replies
6
Views
998
  • Introductory Physics Homework Help
Replies
15
Views
380
  • Introductory Physics Homework Help
Replies
6
Views
1K
Back
Top