Physics
posted by Josh .
A man stands on the roof of a 19.0 mtall building and throws a rock with a velocity of magnitude 24.0 m/s at an angle of 38.0 degrees above the horizontal. You can ignore air resistance.
Calculate the maximum height above the roof reached by the rock.
Got it.... 11.1 m
Calculate the magnitude of the velocity of the rock just before it strikes the ground.
Got it.... 30.8 m/s
Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.
I don't know how to calculate this...
Since you got the first parts, I will just describe how to do the last part.
The horizontal distance travelled will equal the horizontal component of velocity (which remains constant), multipled by the time of flight. You can get the total time by adding the time taken going up (t1 = Vo sin 38/g) to the time it taken going down. The time taken going down, t2, can be derived from the sum of the building height (19 m) and the maximum height reached above the roof (11.1 m)
Total max height above ground = 30.1 m = (1/2) g t2^2
Solve for t2 and than add it to t1 for the time of flight. With that, get the horizontal distance travelled by the method mentioned above
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