Projectile Motion Homework Help

AI Thread Summary
To find the initial velocity and maximum height of a projectile launched at an angle of 28 degrees with a horizontal range of 68 meters and a time of 6.30 seconds, the horizontal component of velocity can be calculated using the equation u(h) = s(h)/t, which gives a constant horizontal velocity. Since there is no horizontal acceleration, the initial horizontal velocity equals the horizontal component of the velocity. To determine the overall initial velocity, trigonometric functions can be applied using the angle provided. For maximum height, focus solely on the vertical components and apply kinematic equations. This approach effectively combines horizontal and vertical motion principles in projectile motion analysis.
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Homework Statement



Find the initial velocity and the maximum height

θ = 28˚
horizontal range/distance/s = 68 m
time = 6.30 s

Homework Equations



suvat equations not so sure which one though

The Attempt at a Solution



I think to find the initial velocity I use the equation:
s=ut+(1/2)at^2

I think acceleration horizontally is 0 so:

u(h) = s(h)/t
 
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loopsnhoops said:

Homework Statement



Find the initial velocity and the maximum height

θ = 28˚
horizontal range/distance/s = 68 m
time = 6.30 s

Homework Equations



suvat equations not so sure which one though

The Attempt at a Solution



I think to find the initial velocity I use the equation:
s=ut+(1/2)at^2

I think acceleration horizontally is 0 so:

u(h) = s(h)/t

There is no acceleration in the horizontal direction so you can simply use s=vt.

Now you can solve for the horizontal component of velocity and you have an angle so using trig you should now be able to solve for the hypotenuse of the initial velocity.

For the maximum height just remember to only focus on the vertical components and use your kinematics equations.
 
loopsnhoops said:

Homework Statement



Find the initial velocity and the maximum height

θ = 28˚
horizontal range/distance/s = 68 m
time = 6.30 s

Homework Equations



suvat equations not so sure which one though

The Attempt at a Solution



I think to find the initial velocity I use the equation:
s=ut+(1/2)at^2

I think acceleration horizontally is 0 so:

u(h) = s(h)/t
Use the last equation, u(h) = s(h)/t to find the horizontal component of the velocity. Since the horizontal component of the velocity is constant, what does this tell you about the horizontal component of the initial velocity?
 
It is the same as the u(h)?
 
Lol I don't know any of this...
 
loopsnhoops said:
It is the same as the u(h)?
Yes. The horizontal component of the velocity doesn't change, so that is the same as the horizontal component of the initial velocity.

From that use some trig to find the initial velocity and the vertical component of the initial velocity.
 
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