Projectile motion problem (find angle theta)

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A projectile is launched from the ground with an initial velocity of 140 m/s at an angle θ, with a time of flight of 18.7 seconds and a range of 1.98 km. To find the launch angle, the horizontal velocity (voh) is calculated as 105.88 m/s using the range and time of flight. The angle θ is then determined using the cosine function, resulting in approximately 41 degrees. The calculations confirm that the launch angle aligns with the given parameters. This demonstrates the application of projectile motion equations effectively.
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Homework Statement


A projectile is launched from the ground with an initial velocity of 140m/s at an angle of θ above the horizontal. (assume AR is negligible and the ground is level)
The time of flight is 18.7seconds and its range is 1.98*10^3
show that the launch angle θ is approximately 41 degrees

Homework Equations


I'm unsure which equation I am supposed to use for a problem like this


The Attempt at a Solution


voh = 140cosθ

∴ cosθ = voh/140

but I am not given enough information to find the initial horizontal velocity so this equation won't work. I also am sure that I have to use the time of flight and the range since its been given but I can't think how to rearrange things to make it work.
Can anyone point me in the right direction for what I'm supposed to be doing please?
 
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5ymmetrica1 said:

Homework Statement


A projectile is launched from the ground with an initial velocity of 140m/s at an angle of θ above the horizontal. (assume AR is negligible and the ground is level)
The time of flight is 18.7seconds and its range is 1.98*10^3
show that the launch angle θ is approximately 41 degrees

Homework Equations


I'm unsure which equation I am supposed to use for a problem like this


The Attempt at a Solution


voh = 140cosθ

∴ cosθ = voh/140

but I am not given enough information to find the initial horizontal velocity so this equation won't work. I also am sure that I have to use the time of flight and the range since its been given but I can't think how to rearrange things to make it work.
Can anyone point me in the right direction for what I'm supposed to be doing please?


What must voh be if the projectile flew for 1.98e3 m and took 18.7sec. to do so?
 
thanks for the reply rudeman!

So voh = 1.8e3m/ 18.7s = 105.88 m/s

then θ = cos-1 (105.88/140)

∴ θ = 40.86 degrees ≈ 41 degrees
 
5ymmetrica1 said:
thanks for the reply rudeman!

So voh = 1.8e3m/ 18.7s = 105.88 m/s

then θ = cos-1 (105.88/140)

∴ θ = 40.86 degrees ≈ 41 degrees

Yay team!
 
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