What Height Will the Projectile Strike the Barrier?

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SUMMARY

The projectile is launched at a speed of 40 m/sec and an angle of 30 degrees, aiming to determine the height at which it strikes a barrier located 85 meters away. The time taken for the projectile to reach the barrier is calculated as approximately 4.08 seconds. The horizontal distance covered is computed using the cosine of the launch angle, resulting in a distance of 22 meters. The vertical displacement is calculated using the equation for motion under constant acceleration, but the initial calculations for vertical displacement were incorrect due to misapplication of acceleration values.

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Homework Statement


A projectile is thrown from the horizontal ground with a speed of 40 m/sec at an angle with the horizontal of 30 degrees. At what height will the projectile strike a high barrier located at a distance of 85m from the firing point?


Homework Equations


ok so: point A is located at the point in which the projectile is launched
point B is located at the point at which the projectile strikes the barrier
point is located at the bottom of the barrier, on the ground directly below the striking point
theta = 30 degrees

change in time from A to B= [VB - VA] / acceleration

dAB = distance from A to C * cos(theta)

dBCx = VB (change in time from B to C) + 0.5a*(change in time B to C)2

The Attempt at a Solution



change in time A to B = [0 - 40] / 9.8 = 4.08 sec

dAB = 85m * cos(30) = 22m

dBCx = 0 (4.08 sec) + 0.5 (9.8) (4.08)2 = 245 m
^my teacher said that this wasn't right. i think i went wrong on the acceleration-i don't think i am calculating it right or using the right value.
 
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You're value for acceleration is fine in the y direction.

How did you derive your equation for time?
 

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