What Height Will the Projectile Strike the Barrier?

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A 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 85m away. The time taken for the projectile to reach the barrier was calculated as approximately 4.08 seconds, using the formula for change in time based on initial and final velocities. The horizontal distance covered was calculated using the cosine of the launch angle, resulting in a distance of 22m. The vertical displacement was calculated incorrectly, leading to confusion about the acceleration value used. Clarification is sought on the derivation of the time equation to ensure accurate calculations.
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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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