Projectiles launched at an angle question

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SUMMARY

The discussion focuses on calculating the maximum height of a projectile launched at an angle of 27.0° with an initial velocity of 20.0 m/s. The vertical component of the initial velocity is determined using the equation v_y = v * sin(θ), resulting in approximately 9.08 m/s. The time to reach maximum height is calculated using the formula t = (v_f - v_i) / a, yielding approximately 1.85 seconds. The kinematic equation d = v_i * t + 0.5 * a * t² is suggested for finding the maximum height, emphasizing the importance of understanding the vertical motion separately from horizontal motion.

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  • Familiarity with trigonometric functions (sine and cosine)
  • Knowledge of kinematic equations
  • Basic grasp of acceleration due to gravity (g = 9.81 m/s²)
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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
 
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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?
 
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 ?
 
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?
 
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.
 

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