Find Distance of Projectile Fired with an Initial Velocity - Help Needed

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To find the distance a projectile travels when fired with an initial vertical velocity of 20 m/s and a horizontal velocity of 30 m/s, the total flight time must first be calculated. The vertical motion is influenced by gravity, allowing the time to reach the peak to be determined as 2 seconds, leading to a total flight time of 4 seconds. The horizontal distance is then calculated by multiplying the horizontal velocity by the total flight time, resulting in a distance of 120 meters. This method effectively combines the principles of projectile motion to arrive at the solution. The final distance from the launch to landing point is 120 meters.
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A projectile is fired over level ground with an initial velocity that has a vertical component of 20 m/s and a horizontal component of 30 m/s. Using g = 10 m/s², find the distance from launching to landing points.



I don't know what formula to use to solve this question. I would appreciate your help.



I start off by finding the magnitude of the resultant vector: √30² + 20², then I don't know what to do next...
 
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Relate the time, velocity, and distance in the x-direction with the time, distance, velocity, and acceleration in the y direction.
 
So this how you do it:
Horizontal velocity multiplied by flight time equals range.

Find the flight time by working out how long it will take for gravity to reduce the initial vertical velocity to zero. This gives you the "time to go up" and this will be equal to the "time to go down".

a=Δv/Δt so if g=-10ms-2 then:
Δt=Δv/g=0-20/-10=2s so total flight time is 4s.

Now just multiply the horizontal velocity by the flight time 30*4=120m

:smile:
 
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