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.
DON'T KNOW
an 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.
DON'T KNOW
You know the initial horizontal velocity (24*cos38). You can figure the time in air (a little tricky, using the vertical distance equation
hinitial=hfinal + 24Sin38*t  1/2 9.8 t^2, solve this quadratic for time, use the quadratic formula)
Now, knowing horizontal velocity and time in air, you can easily calculate horizontal distance.
calculating the horizontal distance is this:
x=V0t * t
so ur initial velocity is 24cos(38) * t
t is the time it took to reach the ground
hi, how did u get the velocity of the rock when it hits the ground?
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